Correct Option: (A)

(1) and (4) only $K E_{\text{avg}\&\text{nbsp};} = \frac{3}{2} k_{B} T$ $P \propto T \left(V = \&\text{nbsp};\text{constant}\&\text{nbsp};\right)$

Correct Option: (A)

$4.14 \times 10^{-15} {~Pa}$ We know that the gas pressure is given as $P = \frac{1}{3} \rho v_{r m s}^{2}$ Now we can write the density of gas as $\rho = m_{0} \&\text{nbsp}; N$, where $m_{0}$ is the mass of the gas molecule and _N_ is the number of molecules in $1 \text{ } m^{3}$ of the gas. In the present case, $N = 100 \times 1 0^{6}$ Or $P = \frac{1}{3} \left(100 \times 1 0^{6}\right) m_{0} v_{r m s}^{2}$ As we know _rms_ velocity of gas molecules is given as $v_{r m s} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 k T}{m_{0}}} \left[A s M = m_{0} N_{a v} \&\text{nbsp};\&\text{nbsp}; a n d \&\text{nbsp}; \frac{R}{N_{A v}} = k\right]$ So we have $P = \frac{1}{3} \times \left(100 \times 1 0^{6}\right) \times 3 k T \left(a s \&\text{nbsp}; T = 3 k\right)$ $P = 4.14 \times 1 0^{- 15} \&\text{nbsp}; P a$

Correct Option: (A)

$\frac{1}{2}k_BT$  Energy associated with each degree of freedom per molecule  $\frac{1}{2} k_{B} T$

Correct Option: (C)

Statement I is true but Statement II is false.  $K_{R} = k T$   $K_{T } = \frac{3}{2} k T$   $K_{T} f o r e a c h m o l e c u l e = \frac{3}{4} k T$  Translational kinetic energy will be equal to rotational kinetic energy corresponds to each degree of freedom So, statement 1 is true and statement 2 is false

Correct Option: (C)

4.0 × 10$^{18}$ Given, height, h = 2cm Kinetic energy, U = 4 × 10<$^{–14}$ erg Density of mercury, ρ = 13.6g / cm<$^{3}$ Volume, V = 4cm<$^{3}$ g = 980 cm/s<$^{2}$ According to gas equation, pV = NKT ⇒ N =  $\frac{p V}{K T}$   For monoatomic gas, Energy,  $U = \frac{3}{2} K T \Rightarrow K T = \frac{2}{3} U$   Now, $N = \frac{3 p V}{2 U}$   So,  $N = \frac{3 g \rho h V}{2 U} \&\text{nbsp};\&\text{nbsp};\&\text{nbsp};\&\text{nbsp};\&\text{nbsp};\&\text{nbsp}; \left(\because Pr e s s u r e , \&\text{nbsp}; p = \rho g h\right) \&\text{nbsp};\&\text{nbsp};\&\text{nbsp};\&\text{nbsp};\&\text{nbsp};\&\text{nbsp};\&\text{nbsp}; \ldots \left(i\right)$  Substituting the given values in equation (i), we get  $N = \frac{3 \times 980 \times 13.6 \times 2 \times 4}{2 \times 4 \times 10^{- 14}} = 3.99 \times 10^{18} \approx 4 \times 10^{18}$  So, correct option is (c).

Correct Option: (B)

6.90 × 10$^{–21}$ Joule/molecule $\begin{matrix}A d i a t o m i c g a s m o l e c u l e p o s s e s s e s t w o r o t a t i o n a l d e g r e e s o f \\ f r e e d o m . \\ H e n c e , t o t a l r o t a t i o n a l k i n e t i c e n e r g y \\ = 2 \frac{1}{2} k T = k T \\ = 1.38 10^{ 23} 500 \\ = 690 10^{ 23} J / m o l e c u l e \\ = 6.90 10^{ 21} J / m o l e c u l e\end{matrix}$

Correct Option: (D)

$\begin{matrix}M e a n K i n e t i c E n e r g y = \frac{3}{2} P V \\ 1.5 10^{3} = \frac{3}{2} P \left(20\right) \\ P = \frac{2}{3} \frac{1.5 10^{3}}{20} = 50 N / m^{2}\end{matrix}$

Correct Option: (A)

50g

Correct Option: (D)

10 x 10$^{–14}$ _erg_ $\begin{matrix}\begin{matrix}A v e r a g e k i n e t i c e n e r g y\end{matrix} \\ \begin{matrix}E = \frac{3}{2} k T \frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}} = \frac{\left(273 23\right)}{\left(273 + 227\right)} = \frac{250}{500} = \frac{1}{2}\end{matrix} \\ E_{2} = 2 E_{1} = 2 5 1 0^{ 14} = 10 1 0^{ 14} e r g\end{matrix}$

Correct Option: (D)

${\displaystyle \left(\frac{7}{88}\right)\mathrm{N}\mathrm{k}\mathrm{T}}$  $\begin{matrix}\begin{matrix}\begin{matrix}M e a n k i n e t i c e n e r g y f o r n m o l e g a s = n . \frac{f}{2} R T\end{matrix}\end{matrix} \\ \begin{matrix}\begin{matrix} E = n \frac{7}{2} R T = \left(\frac{m}{M}\right) \frac{7}{2} N k T = \frac{1}{44} \left(\frac{7}{2}\right) N k T\end{matrix}\end{matrix} \\ \begin{matrix}\begin{matrix}= \frac{7}{88} N k T \left[A s f = 7 a n d M = 44 f o r C O_{2}\right]\end{matrix}\end{matrix}\end{matrix}$

Correct Option: (C)

the temperature of the gas only. Kinetic energy ∝ Temperature

Correct Option: (D)

3.0 _atm_ $v_{r m s} = \sqrt{\frac{3 P}{\rho}} \Rightarrow P = \frac{v_{r m s}^{2} \times \rho}{3} = \frac{\left(3180\right)^{2} \times 8.99 \times 10^{- 2}}{3}$ = 3.03 × 10<$^{5}$ _N__/m_<$^{2}$ = 3 atm.

Correct Option: (D)

7.7× 10$^{$^{3}$}$ K $\begin{matrix}\frac{3}{2} k T = 1 e V T = \frac{2}{3} \frac{e V}{k} = \frac{\frac{2}{3} 1.6 10^{ 19}}{1.38 10^{ 23}} \\ = 7.7 10^{3} K\end{matrix}$

Correct Option: (A)

1 : 2

Correct Option: (D)

3.4× 10$^{3}$ _J_ $\begin{matrix}K i n e t i c e n e r g y p e r g m o l e E = \frac{f}{2} R T \\ I f n o t h i n g i s s a i d a b o u t g a s , t h e n w e s h o u l d c a l c u l a t e t h e t r a n s l a t i o n a l \\ k i n e t i c e n e r g y \\ \begin{matrix}i . e . , E_{T r a n s} = \frac{3}{2} R T = \frac{3}{2} 8 .31 \left(273 + 0\right)\end{matrix} \\ \begin{matrix}= 3 . 4 1 0^{3} J\end{matrix}\end{matrix}$

Correct Option: (C)

1.01 x 10$^{5}$ _N_/_m_$^{2}$ $\Rightarrow P = \frac{\rho V_{r m s}^{2}}{3}$ By substituting the given values, we get $= \frac{8.99 \times 10^{- 2} \times 1840 \times 1840}{3} = 1.01 \times 1 0^{5} \text{ } N / m^{2}$

Correct Option: (A)

uniform everywhere. When gas is filled in a closed container, it exerts pressure on the walls of the vessel. According to kinetic theory, this pressure is developed due to the collisions of the moving molecules on the walls of the vessels. Whenever a molecule collides with the wall, it returns with changed momentum, and an equal momentum is transferred to the wall. According to Newton’s law of motion, the rate of change of momentum of the ball is equal to the force exerted on the wall. Since the gas contains many molecules that are colliding with the walls of the vessel, they exert a steady force on the walls. This force measured per unit area gives pressure, which is the same as the molecules are moving in the horizontal direction with constant acceleration.

Correct Option: (A)

(V$_{rms}$)$_{1}$ < (V$_{rms}$)$_{2}$ < (V$_{rms}$)$_{3}$ and (K.E)$_{1}$ = (K.E)$_{2}$ = (K.E$_{3}$) $\begin{matrix}v_{r m s} \frac{1}{\sqrt{M}} \left(v_{r m s}\right)_{1} < \left(v_{r m s}\right)_{2} < \left(v_{r m s}\right)_{3} \\ a l s o i n m i x t u r e t e m p r a t u r e o f e a c h g a s w i l l b e s a m e h e n c e k i n e t i c \\ e n e r g y a l s o r e m a i n s s a m e .\end{matrix}$

Correct Option: (A)

2490 _joules_ $\begin{matrix}K i n e t i c E n e r g y \left(K . E\right) = \frac{3}{2} n R T \\ G i v e n , T = 47 ^{} C \\ = 47 + 273 = 320 K \\ N o w , \\ K E = \frac{3}{2} n R T \\ = \frac{3}{2} \left(\frac{m}{M}\right) R T \left( n = \frac{m}{M}\right) \\ = \frac{3}{2} \left(\frac{20}{32}\right) 8.3 320 \\ K E = 2490 J\end{matrix}$

Correct Option: (C)

doubled. Kinetic energy ∝ Temperature. Hence if the temperature is doubled, kinetic energy will also be doubled.

Correct Option: (D)

${\displaystyle \frac{1}{2}\phantom{\rule{\;}{0ex}}RT}$  $\begin{matrix}T h e m a i n k i n e t i c e n e r g y o f o n e m o l e o f g a s n d e g r e e o f f r e e d o m \\ E = \frac{n}{2} R T \\ T h e m e a n k i n e t i c e n e r g y o f o n e m o l e o f g a s p e r d e g r e e o f f r e e d o m \\ E^{'} = \frac{E}{n} = \frac{\frac{n}{2} R T}{n} \\ E^{'} = \frac{1}{2} R T\end{matrix}$

Correct Option: (B)

1.11 × 10$^{32}$ $\begin{matrix}W e h a v e b e e n t o l d t h a t n u m b e r o f c o l l i s i o n s p e r s e c o n d i s t h e r e c i p r o c a l o f \\ t h e c o l l i s i o n t i m e t h e r e f o r e , w e n e e d t o f i n d t h e t i m e o f c o l l i s i o n \\ T \left(T h e m e a n f r e e t i m e\right) . \\ T = \frac{k_{B} T}{4 \sqrt{2} r^{2} P v_{e m s}} \left(M e a n f r e e t i m e l a w\right) \\ T = \frac{k_{B} T}{4 \sqrt{2} r^{2} P v_{r m s}} \\ N o t e t h a t , v_{r m s} = \sqrt{\frac{3 R T}{M}} \\ N o t e t h a t , t h e g i v e n g a s i s t h e n i t r o g e n v_{r m s} = \sqrt{\frac{3 R T}{M_{N_{2}}}} . . . . . \left(2\right) \\ N o t e t h a t , M_{N_{2}} = 2 M_{N} = 2 14.01 = 28.02 g / m o l \\ M_{N_{2}} = 28.02 10^{ 3} k g / m o l \\ P u t t h e v a l u e o f M_{N_{2}} f r o m \left(3\right) i n t o \left(2\right) , \\ v_{r m s} = \sqrt{\frac{3 8.31 273}{28.02 10^{ 3}}} , v_{r m s} = 493 m / s . . . . \left(4\right) \\ P u t t h e v a l u e o f v_{r m s} f r o m \left(4\right) i n t o \left(1\right) , \\ T = \frac{1.38 10^{ 23} 273}{4 \sqrt{2} \left[1.88 10^{ 10}\right]^{2} 1.01 10^{5} 493} , T = 1.205 10^{ 10} s . . . . . \left(5\right) \\ N o w w e c a n f i n d t h e n u m b e r o f c o l l i s i o n s i n 1 .0 s b y g e t t i n g t h e r e c i p r o c a l \\ o f t h e c o l l i s i o n t i m e , \end{matrix}$ $\begin{matrix}N_{c o l l i s i o n s} = \frac{ t}{} = \frac{1.0}{} \\ N_{c o l l i s i o n s} = ^{ 1} = \left[1.205 10^{ 10}\right]^{ 1} \\ N_{c o l l i s i o n s} = 8.30 10^{9} c o l l i s i o n . . . . . \left(6\right) \\ N o t e t h a t , t h e v a l u e o f t h e p r e c e d i n g e q u a t i o n i s t h e n u m b e r o f c o l l i s i o n s f o r \\ o n e m o l e c u l e i n o n e s e c o n d . S o , t o f i n d t h e t o t a l n u m b e r o f c o l l i s i o n s w e n e e d \\ t o f i n d t h e t o t a l n u m b e r o f m o l e c u l e s . \\ T o f i n d t h e n u m b e r o f n i t r o g e n m o l e c u l e s i n 1 .0 L a t t e m p e r a t u r e o f 273 K \\ a n d a p r e s s u r e o f 1.0 a t m , \\ P V = N k_{B} T \left(I d e a l g a s l a w\right) \\ N = \frac{P V}{k_{B} T} \\ N = \frac{1.01 10^{5} 10^{ 3}}{1.38 10^{ 23} 273} \\ N = 2.681 10^{22} m o l e c u l e . . . . . . \left(7\right) \\ N o t e t h a t , e a c h c o l l i s i o n i n v o l v e s t w o m o l e c u l e s \\ w e a r e s u p p o s i n g t h a t e v e r y t w o m o l e c u l e s w i l l c o l l i d e , a n d w e w i l l n e g l e c t t h e \\ c o l l i s i o n s w i t h t h e w a l l s o f t h e c o n t a i n e r . \\ T h e r e f o r e , w e w i l l d i v i d e t h e n u m b e r o f m o l e c u l e s b y 2 . T h i s w i l l g i v e u s t h e \\ n u m b e r o f c o l l i s i o n s s i m u l t a n e o u s l y i n a n i n f i n i t e s i m a l t i m e , W e w i l l m u l t i p l y \\ t h e r e s u l t b y t h e n u m b e r o f c o l l i s i o n s i n o n e s e c o n d \left[e q u a t i o n \left(6\right)\right] . \\ N_{n e t , c o l l i s i o n s} = \frac{N}{2} N_{c o l l i s i o n s} \left(N f r o m \left(7\right) , a n d N_{c o l l i s i o n s } f r o m \left(6\right)\right) \\ N o t e t h a t , e a c h c o l l i s o n i n v o l v e d t w o m o l e c u l e s , t h a t^{'} s w h y w e h a n d t o d i v i d e t h e \\ n u m b e r o f m o l e c u l e s b y 2. \\ N_{n e t , c o l l i s i o n s} = \frac{2.681 10^{22}}{2} 8.30 10^{9} \\ N_{n e t , c o l l i s i o n s} = 1.113 10^{32} C o l l i s i o n s\end{matrix}$

Correct Option: (A)

137°C $\begin{matrix}A v e r a g e k i n e t i c t h e o r y o f o n e m o l e c u l e i s E = \frac{3}{2} k T \\ w h e r e i s B o l t z m a n n c o n s t a n t a n d t h e a b s o l u t e t e m p e r a t u r e . \\ G i v e n , T_{1} = 68^{} C = 273 68 = 205 K , \\ E_{1} = E , E_{2} = 2 E \\ \frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}} \\ T_{2} = \frac{T_{1} E_{2}}{E_{1}} \\ T_{2} = \frac{205 2 E}{E} = 410 K\end{matrix}$ $T_{2} = 410 273 = 13 7^{0} C$

Correct Option: (D)

35 _K$_{B}$T_ Degree of freedom(_f_) = 5 + 2(3_N_ – 5)                                     _f_ = 5 + 2(3 × 2 –1) = 7 energy of one molecule =$\textrm{ }\textrm{ } \frac{f}{2} K_{B} T$ energy of 10 molecules =

Correct Option: (C)

$\frac{1}{\sqrt{2} n \pi d^2}$ n = number of molecule per unit volume d = diameter of the molecule $\lambda = \frac{1}{\sqrt{2} \pi \left(\&\text{nbsp}; d\right)^{2} n} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ }$(By Theory)

Correct Option: (A)

$V_{m s} = \sqrt{\frac{3 R T}{M}}$ $V_{r m s}^{2} = 3 R T / M$ Hence we can conclude that $V_{r m s}^{2}$ is directly proportional to temperature y = mx ⇒ Graph will be straight line

Correct Answer: 1

Average rotation kinetic energy per molecule $= \frac{f}{2} k T = \frac{2}{2} k T = k T$ $\frac{\left(K . E\right)_{O_{2}}}{\left(K . E\right)_{N_{2}}} = \frac{k T}{k T} = 1.$

Correct Answer: 313

_T_<$_{1}$ = 20°_C_ = 273 + 20 = 293 _K_ _$E_{1} = \frac{3}{2} k T_{1} \text{ } a n d \text{ } E_{2} = \frac{3}{2} k T_{2}$_ __E_<$_{2}$ = 2_E_<$_{1}$_ $\frac{3}{2} k T_{2} = 2 \times \frac{3}{2} k T_{1}$ _T_<$_{2}$ = 2_T_<$_{1}$ = 2 × 293 = 586 _K_ _T_<$_{2}$ = 586 – 273 = 313°_C_

Correct Answer: 2

$P_{0} = \frac{m n V_{r m s}^{2}}{3 V}$ $I f \text{ } m^{1} \text{ } = \frac{m}{2} \text{ } \text{ } a n d \text{ } \text{ } V_{r m s}^{1} \text{ } = \text{ } 2 V_{r m s}$ $P^{'} = 2 P 0$

Correct Option: (D)

1 : 1 Average kinetic energy per molecule is given as $K_{a v} = \frac{f}{2} k_{B} T$ $\frac{\left(K_{a v}\right)_{H_{2}}}{\left(K_{a v}\right)_{O_{2}}} = \frac{f_{H_{2}}}{f_{O_{2}}}$ Ratio = 1 : 1

Correct Option: (B)

$\frac{3 P_0 V_0}{4}$ $P = \frac{P}{1 + \left(\frac{V}{V_{0}}\right)^{3}} = \frac{P_{0}}{2}$ $T = \frac{P_{0} V_{0}}{2 R}$ Therefore, translational kinetic energy is equal to $\frac{3}{2} R T = \frac{3 R}{2} \frac{P_{0} V_{0}}{2 R} = \frac{3 P_{0} V_{0}}{4}$

Correct Option: (D)

$\frac{3 R T}{2 E}$ Average kinetic energy per unit molecule $E = \frac{3}{2} k T$ $\therefore k = \frac{2 E}{3 T}$ But Avogadro number $= N_{A} = \frac{R}{k} = \frac{R}{\left(\frac{2 E}{3 T}\right)}$ $\therefore N_{A} = \frac{3 R T}{2 E} .$

Correct Option: (A)

Less than 1 _eV_ = 6.27 × 10<$^{–21}$ _J_ = 0.039 _eV_ < 1 _eV_

Correct Option: (C)

remains unchanged. According to kinetic theory, molecules of a liquid are in a state of continuous random motion. They continuously collide against the walls of the container. During each collision, certain momentum is transferred to the walls of the container. So, kinetic energy of molecules increases, hence due to random motion, the temperature increase. So, random motion of molecules and not ordered motion cause rise of temperature.

Correct Option: (D)

6167.831 J Given values from the question: Mass of $\text{CO}_{2}$ = 75 g Temperature = $17^{\circ}\text{C}$ We will also use the following constants: Molar mass of $\text{CO}_{2}$ (M) = 44 g/mol Universal gas constant (R) = 8.314 J/(mol·K) The formula for the translational kinetic energy of a gas is given by: $ KE_{\text{trans}} = \frac{3}{2} nRT\!$ where 'n' is the number of moles and 'T' is the absolute temperature in Kelvin. Step 1: Convert the temperature from Celsius to Kelvin. $T(K)=T(^{\circ}\text{C})+273.15=17+273.15=290.15\!$ Step 2: Calculate the number of moles (n) of $\text{CO}_{2}$. Using the formula for 1.705: $n=\frac{\text{Given Mass}}{\text{Molar Mass}}=\frac{75}{44}=1.705\!$ The number of moles is 1.705. Step 3: Calculate the translational kinetic energy. Now, substitute all the values into the kinetic energy formula : $KE_{\text{trans}}=\frac{3}{2}\ \times\ 1.705\ \times\ 8.314\ \times\ 290.15=6167.831\ J\!$ The calculated translational kinetic energy is 6167.831 J.

Correct Option: (B)

$315 ^{\circ} C$ The average kinetic energy (KE) of gas molecules is directly proportional to the absolute temperature in Kelvin. i.e. $ KE \propto T\!$ This means that to double the kinetic energy, the absolute temperature must also be doubled. Given the initial temperature: $T_{initial}=21^{\circ}\text{C}\!$ Step 1: Convert the initial temperature from Celsius to Kelvin. Using : $T(K)=T(^{\circ}\text{C})+273\!$ $\Rightarrow T_{initial}(K)=21+273=294\ \!$ The initial temperature is 294 K. Step 2: Calculate the final absolute temperature required to double the kinetic energy. $\Rightarrow T_{final}(K)=2\ \times\ 294=588\!$ The final temperature in Kelvin is 588 K. Step 3: Convert the final temperature from Kelvin back to Celsius. $\Rightarrow T(^{\circ}\text{C})=T(K)-273=588-273=315\!$ The final temperature in Celsius is $315 ^{\circ} C$ .

Correct Option: (C)

$3315.942\ \text{ K}\!$ Given values: $E=0.429\text{ eV}=0.429\ \times\ 1.6\ \times\ 10^{-19}\text{ J}\!$ $K_B=1.38\ \times\ 10^{-23}\text{ J/K}\!$ The formula for the average kinetic energy of a monoatomic molecule is: $ E = \frac{3}{2} K_B T\!$ Now, rearrange the kinetic energy formula to solve for the temperature (T): $\Rightarrow T=\frac{2E}{3K_B}=\frac{2\ \times\ \left(0.429\ \times\ 1.6\ \times\ 10^{-19}\right)\ }{3\ \times\ 1.38\ \times\ 10^{-23}}\!$ $\Rightarrow T=3315.942\ \text{ K}\!$ Therefore, the temperature is 3315.942 K.

Correct Option: (C)

11RT $U = n_{1} C_{v 1} T + n_{2} C_{v 2} T$ $= \frac{R}{\gamma_{1} - 1} T + \frac{4 R}{\gamma_{2} - 1} T$ $= \frac{2 R}{\frac{7}{5} - 1} T + \frac{4 R}{\frac{5}{3} - 1} T$ $= R T \left[5 + 6\right]$ $U = 11 R T$

Correct Option: (A)

$K E = \frac{1}{2} m V_{a v g}^{2}$ $k E = \frac{1}{2} m \left(\sqrt{\frac{8 R T}{\pi m}}\right)^{2}$ $k E = \frac{4 R T}{\pi}$ $k E = \frac{4 R T}{\pi}$only depends on temperature. $\frac{K E_{H_{2}}}{K E_{A r}} = 1$

Correct Option: (C)

$V _{ rms }({\rm mono })> V _{ rms }({\rm dia })> V _{ rms } ({\rm poly})$ $V_{r m s \text{ }} \alpha \frac{1}{\sqrt{M}}$ $M_{P o l y a t o m i c} > M_{D i a t o m i c} > M_{m o n o a t o m i c}$ $\Rightarrow V_{r m s \left(m o n o\right)} > V_{r m s \left(D i a\right)} > V_{r m s \left(p o l y\right)}$

Correct Option: (A)

1.33 The kinetic energy of gas,  $E = \frac{3}{2} n R T \Rightarrow E \propto T$ $\therefore \frac{E ^{'}}{E} = \frac{T ^{'}}{T} = \frac{400}{300}$ $\Rightarrow \frac{E ^{'}}{E} = \frac{4}{3} = 1.33$

Correct Option: (D)

$P=\frac{2}{3} E$ From Kinetic Theory of Gases. $K \cdot E = \frac{3}{2} R T$ From the Ideal Gas Equation _P V_ = _R T_  ( for 1 mole ) Substituting the value of  _R T_  from the above equations of the kinetic theory of gases we get, $\frac{K \cdot E}{V} = \frac{3}{2} P$ $\Rightarrow P = \frac{2}{3} \left(\frac{K \cdot E}{V}\right) = \frac{2}{3} E$

Correct Option: (D)

4 $v_{\text{avg}\&\text{nbsp};} \propto \sqrt{T}$ $\frac{\left(V_{\text{avg}\&\text{nbsp};}\right)_{2}}{\left(V_{\text{avg}\&\text{nbsp};}\right)_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$ $\frac{2 \cdot \left(V_{\text{avg}\&\text{nbsp};}\right)_{1}}{\left(V_{\text{avg}\&\text{nbsp};}\right)_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$ $2 = \sqrt{\frac{T_{2}}{T_{1}}}$ $\Rightarrow T_{2} = 4 T_{1}$ From _P V_ = _n R T_  Closed container ⇒ V = Constant _P_ ∝ _T_  $\frac{P_{2}}{P_{1}} = \frac{T_{2}}{T_{1}}$ $P_{2} = P_{1} \left(\frac{T_{2}}{T_{1}}\right) = P_{1} \left(\frac{4 T_{1}}{T_{1}}\right)$ $P_{2} = 4 P_{1}$

Correct Option: (B)

2049 d Mean Free path $\left(\overset{ˉ}{\lambda}\right) = \frac{1}{\sqrt{2} \pi n d^{2}} = \frac{K_{B} T}{\sqrt{2} \pi d^{2} P}$ Clearly $\overset{ˉ}{\lambda} \propto T$ $\frac{\overset{\overline}{\lambda^{'}}}{\overset{ˉ}{\lambda}} = \frac{373}{273}$ $\overset{ˉ}{\lambda} ^{'} = \frac{373}{273} \times \overset{ˉ}{\lambda} = \frac{373}{273} \times 1500 \textrm{ }\textrm{ } d = 2049.45 \textrm{ }\textrm{ } d$

Correct Option: (C)

651.7 m/s $V_{r m s} \times \frac{1}{\sqrt{M}}$  $\Rightarrow \frac{V_{\text{rms}\&\text{nbsp};} \&\text{nbsp};\text{of}\&\text{nbsp};\text{Ar}\&\text{nbsp};}{490} = \sqrt{\frac{70.9}{39.9}}$ $\Rightarrow V_{r m s} o f \text{ } A r g o n = 490 \times 1.3 = 651.7 \text{ } m / s$

Correct Option: (D)

327$^{o}$C $\Rightarrow 600 = 273 + t$ $\Rightarrow t = 600 - 273 = 32 7^{\circ} C$

Correct Option: (D)

$2 v _0$ $v_{r m s} = \sqrt{\frac{3 R T}{M}}$ $\begin{matrix}\frac{v^{'}}{v_{0}} = \sqrt{\frac{800}{200}} = 2 \\ \Rightarrow v^{'} = 2 v_{0}\end{matrix}$

Correct Option: (A)

0.25 Collision frequency, $f = \frac{v}{\lambda}$  $\&\text{nbsp};\text{or}\&\text{nbsp}; f = \frac{v}{\left(\frac{1}{\sqrt{2} \pi d^{2} n_{v}}\right)} = \sqrt{2} \pi d^{2} v n_{v}$ $\text{ } \therefore f \propto n_{v} \text{ }$, where $\text{ } n_{v} =$ no. density or $\text{ } \frac{f_{1}}{f_{2}} = \frac{n_{v_{1}}}{n_{v_{2}}} \Rightarrow \frac{f_{1}}{f_{2}} = \frac{3 \times 1 0^{19}}{12 \times 1 0^{19}} = \frac{1}{4} = 0.25$

Correct Answer: 51

Correct Option: (B)

4102.8 _J_ $\left(\right. K E \left.\right)_{\text{Translational}\&\text{nbsp};} = \left[\frac{3}{2} K T\right] \times$ no. of molecule No. of molecule $= \left[\frac{50}{44} \times 6.023 \times 1 0^{23}\right]$ $\left(\right. K E \left.\right)_{\text{Translational}\&\text{nbsp};} = 4108.644 \&\text{nbsp}; J$

Correct Option: (C)

$\frac{5}{2}$ $\frac{\rho_{1}}{\rho_{2}} = \frac{4}{25}$ Ratio of rms velocities $= \sqrt{\frac{\rho_{2}}{\rho_{1}}} = \frac{5}{2}$

Correct Option: (D)

1 : 1 By the equipartition of energy per molecule where $k_{B}$ is the Boltzmann's constant. Hence, for the same given temperature $\left(T = 300 K\right)$ $K . E . \propto f$ (degree of freedom) For He and Ar _f_ = 3 $\frac{K \cdot E_{H e}}{\text{ } K \cdot E_{A r}} = \frac{1}{1}$

Correct Option: (B)

$P_2=P_1,\; T_2=\frac{T_1}{2}$ $E = \frac{3}{2} N k T$  Initially $E_{1} = \frac{3}{2} N_{1} k T_{1}$  and finally $E_{2} = \frac{3}{2} N_{2} k T_{2}$  But according to the problem _E_<$_{1}$ = _E_<$_{2}$ and _N_<$_{2}$ = 2_N_<$_{1}$  $\frac{3}{2} N_{1} k T_{1} = \frac{3}{2} \left(2 N_{1}\right) k T_{2}$  $\Rightarrow T_{2} = \frac{T_{1}}{2}$  Since the kinetic energy is constant $\frac{3}{2} N_{1} k T_{1} = \frac{3}{2} N_{2} k T$  $\Rightarrow N_{1} T_{2} = N_{2} T_{2}$  ⸫ _NT_ = constant  From ideal gas equation of _N_ molecules _PV_ = _NkT_ $\Rightarrow P_{1} V_{1} = P_{2} V_{2}$  $\therefore P_{1} = P_{2}$  (as _V_<$_{1}$ = _V_<$_{2}$ and _NT_ = constant)

Correct Option: (C)

_E_ As the kinetic energy is given by  $K . E . = \frac{3}{2} k_{B} T ;$  _k<$_{B}$_ = Boltzmann constant T = Temperature The kinetic energy per molecule of the gas depends only upon the temperature, therefore kinetic energy will remain the same i.e. _E_.

Correct Option: (A)

426.3 _K_ Given that _V_<$_{rms}$(_O_<$_{2}$) = _V_<$_{rms}$(_N_<$_{2}$) $\Rightarrow \sqrt{\frac{3 R T_{1}}{M_{1}}} = \sqrt{\frac{3 R T_{2}}{M_{2}}}$  Where _M_<$_{1}$ is Molar mass of _O_<$_{2}$ = 32 and _M_<$_{2}$ is Molar mass of _N_<$_{2}$ = 28 thus _O_<$_{2}$ is at _T_<$_{1}$ temperature ⇒ _N_<$_{2}$ is at _T_<$_{2}$ = 373 _K_ temperature $T_{1} = \frac{M_{1}}{M_{2}} \times T_{2}$  $T_{1} = \frac{32}{28} \times 373$  _T_<$_{1}$ = 426.3 _K_

Correct Option: (C)

$2P_0$ $P_{0} = \frac{1}{3} \frac{N m}{V} v_{r m s}^{2}$  (_m_ = mass of each molecule, _V_ = volume of the gas) $\text{Now} , m^{'} = \frac{1}{2} m$  $V_{r m s}^{^{'}} = 2 v_{r m s}$  $P^{'} = \frac{1}{3} \frac{N m . 4 v_{r m s}^{2}}{2 V} = 2. \frac{1}{3} \frac{N m}{V} v_{r m s}^{2} = 2 P_{0}$

Correct Option: (D)

$\frac{n_1 T_1+n_2 T_2}{n_1+n_2}$ Since there is no loss of energy  ∴ Sum of change of internal energies must be zero i.e. ∆_U_<$_{1}$ + ∆_U_<$_{2}$ = 0 $\Rightarrow n_{1} C_{v} \left[T - T_{1}\right] + n_{2} C_{v} \left[T - T_{2}\right] = 0$ $\Rightarrow \left(n_{1} + n_{2}\right) T = n_{1} T_{1} + n_{2} T_{2}$ $\Rightarrow T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

Correct Option: (C)

doubled

Correct Option: (A)

its molecular kinetic energy increases The internal energy of a gas is totally kinetic and this kinetic energy (translational) for a molecule of a gas is given by $\text{ } K = \frac{3}{2} k T$ Hence,$K \propto T$ As the temperature of the gas is increased, its molecular kinetic energy increases.

Correct Option: (A)

$14 \times 10^{-13}$ Average KE$\text{ } = \frac{1}{2} m v_{r m s}^{2} = \frac{1}{2} m \cdot \frac{3 P}{\rho}$ $= \frac{1}{2} \frac{M}{N} \times \frac{3 P}{\rho}$ $= \frac{1}{2} \times \frac{28}{6 \times 1 0^{23}} \times \frac{3 \times 1 0^{6}}{\left(\frac{1}{2 \times 1 0^{4}}\right)} = 14 \times 1 0^{- 13} e r g$

Correct Option: (B)

$v_{ {rms }}>v_{ {av }}>v_{ {mp }}$ $v_{r m s} = \sqrt{\frac{3 k T}{m}} , \text{ } v_{a v} = \sqrt{\frac{8 k T}{m \pi}}$ and$v_{m p} = \sqrt{\frac{2 k T}{m}}$ $\therefore v_{r m s} > v_{a v} > v_{m p}$

Correct Option: (B)

proportional to absolute temperature Average kinetic energy of a molecule of gas is given as $\frac{f}{2} K_{B} T$ where _f_ is degree of freedom. which is independent of pressure and volume and depend on the absolute temperature

Correct Option: (C)

Proportional to square root of temperature $\left( {\sqrt T } \right)$ $v_{r m s} = \sqrt{\frac{3 R T}{M}}$

Correct Option: (C)

Mean kinetic energy of gas molecule $E = \frac{f}{2} k T = \frac{f}{2} k \left(t + 273\right) = \left(\frac{f}{2} k\right) t + \frac{f}{2} 273 \text{ } k ;$ Comparing it with standard equation of straight $l i n e \text{ } \&\text{nbsp}; y = m x + c . \text{ }$ $W e \&\text{nbsp}; g e t \text{ } m = \frac{f}{2} k \text{ } a n d \text{ } c \text{ } = \frac{f}{2} 273 \text{ } k$ So, the graph between and will be straight line with positive intercept on E axis and positive slope with tax is.

Correct Option: (C)

0.5 $V_{R M S} \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{V_{O_{2}}}{V_{H_{2}}} = \sqrt{\frac{M_{H_{2}}}{M_{O_{2}}}} \Rightarrow \frac{V_{O_{2}}}{2} = \sqrt{\frac{2}{32}}$ $\Rightarrow V_{O_{2}} = \frac{2}{4} = 0.5 \&\text{nbsp}; k m / s$

Correct Option: (C)

kinetic energy

Correct Option: (B)

20 _K_ **RMS velocity of a gas molecule is given by:** ****$V_{r m s} = \sqrt{\frac{3 R T}{M}}$ Let the temperature at which the velocity of hydrogen is equal to the velocity of oxygen at  $47 \&\text{nbsp}; ° C$   be T  $\sqrt{\frac{3 R T}{2}} = \sqrt{\frac{3 R \left(273 + 47\right)}{32}}$    $\therefore T / 2 = \frac{\left(273 + 47\right)}{32}$ $\Rightarrow T = 20 \&\text{nbsp}; K$

Correct Option: (A)

6232.5 _J_ Since Oxygen is diatomic Kinetic energy $= \frac{5}{2} \text{ } R T$ $= \frac{5}{2} \times 8.31 \times 300$ = 6232.5 _J_

Correct Option: (B)

3200 _K_ $E = \frac{3}{2} K_{B} T$ $0.414 \times 1.6 \times 1 0^{- 19} = \frac{3}{2} \times 1.38 \times 1 0^{- 23} T$ $T = 3200 \&\text{nbsp}; K$

Correct Option: (A)

the force causing the molecules to move across the tube is Δnk$_{B}$TS Correct Answer is:* force balance implies n$_{1}$βvl = Δnk$_{B}$T Correct Answer is: total number of molecules going across the tube per sec is $\left( {\frac{{{\rm{\Delta }}n}}{l}} \right)\left( {\frac{{{k_B}T}}{\beta }} \right){\rm{S}}$ $\therefore P_{1} = n_{1} k_{B} T , P_{2} = n_{2} k_{B} T$ $\therefore \Delta F = \left(P_{1} - P_{2}\right) S$ $= \left(n_{1} - n_{2}\right) k_{B} T S$ $\Delta F = \Delta n_{B} T S$                     ...(i) ∴ Balancing of force, $\beta v \times \left(S I \times n_{1}\right) = \Delta \left(n k\right)_{B} T S$ $\Rightarrow n_{1} \beta v l = \Delta \left(n k\right)_{B} T$             ...(ii) $\therefore v = \frac{\Delta \left(n k\right)_{B} T}{\beta n_{1} I}$ ∴ Number of molecules per second $\frac{\Delta \mathbf{N}}{\Delta t} = \frac{\left(v d t\right) \times \mathbf{S} \times \mathbf{n}_{1}}{d t}$ $= v S n_{1}$ $= \frac{\Delta \left(n k\right)_{B} T}{\beta n_{1} I} \times \left(\mathbf{S} \mathbf{n}\right)_{1}$ $\frac{\Delta N}{\Delta t} = \left(\frac{\Delta n}{I}\right) \left(\frac{k_{B} T}{\beta}\right) S$      ...(iii) ∴ Δ_n_ decreases with time, so rate of transfer decreases with time. ∴ Answer (A, B, C)

Correct Option: (C)

$v_{\mathrm{p}}<\bar{v}

Correct Option: (A)

1 : 1 Average rotational KE $= 2 \times \frac{1}{2} k T$ (for diatomic gas) (f)

Correct Option: (D)

20 K $v_{r m s} = \sqrt{\frac{3 R T}{M}}$ Where T is the temperature of the gas molecules in kelvin and M is the molecular mass of the gas $v_{H_{2}} = v_{O_{2}} \Rightarrow \sqrt{\frac{T_{H_{2}}}{M_{H_{2}}}} = \sqrt{\frac{T_{O_{2}}}{M_{O_{2}}}} \Rightarrow \frac{T_{H_{2}}}{2} = \frac{320}{32}$ $\Rightarrow T_{H_{2}} = 20 \&\text{nbsp}; K$

Correct Option: (C)

remains same The temperature of gas molecules depends on the average kinetic energy associated with the disorderly motion (i.e., random motion) of the gas molecules. The orderly kinetic energy of the molecules of the gas container will increase in the lorry, whereas disorderly kinetic energy will still remain the same; hence the temperature of the gas molecules will remain unchanged.

Correct Option: (C)

$\rm \mathrm{v}_{\mathrm{p}}<\overline{\mathrm{v}}<\mathrm{v}_{\mathrm{rms}}$ Correct Answer is: The average kinetic energy of a molecule is$\rm \;\frac{3}{4} \mathrm{mv}_{\mathrm{p}}^2$ $v_{P} = \sqrt{\frac{2 R T}{M}} ; \overset{\overline}{v} = \sqrt{\frac{8 R T}{\pi M}} ; v_{r m s} = \sqrt{\frac{3 R T}{M}}$

Correct Option: (D)

$\rm 3 \pi / 8$ $\left(v_{m s}\right)_{L} = \left(v_{\text{avg}\&\text{nbsp};}\right)_{R} \sqrt{\frac{3 R T}{M_{1}}} = \sqrt{\frac{8 R T}{\pi M_{2}}} \Rightarrow \frac{M_{1}}{M_{2}} = \frac{3 \pi}{8}$

Correct Option: (B)

The gas possess equal energies in all the three directions x,y and z-axis

Correct Option: (D)

(iii) and (iv) only

Correct Option: (C)

9360 _J_ $\textrm{ }\textrm{ } v_{r m s \textrm{ }\textrm{ }} \propto \sqrt{T}$ $v_{r m s} \propto \sqrt{300 \textrm{ }\textrm{ } K} , v_{r m s_{f}} = 2 v_{r m s_{i}}$ $v_{r m s_{f}} \propto \sqrt{1200 \textrm{ }\textrm{ } K}$ $\textrm{ }\textrm{ } T_{f} = 1200 \textrm{ }\textrm{ } K , \textrm{ }\textrm{ } T_{i} = 300 \textrm{ }\textrm{ } K , n = \frac{14}{28} = \frac{1}{2}$ $Q = n C_{v} \Delta T = \frac{1}{2} \times \frac{5 R}{2} \times 900$ $Q = 9360 \textrm{ }\textrm{ } J$

Correct Answer: 7

In an adiabatic expansion, रुद्धोष्म प्रसार esa TV$^{g–1}$ = constant नियतांक $\begin{matrix}\text{T}_{0}   \text{V}^{^{\gamma - 1}} = \text{T} \left(\frac{\text{V}}{5}\right)^{\gamma - 1} \\ \therefore \gamma = 1 + \frac{2}{5} \\ \Rightarrow \gamma - 1 = \frac{2}{5} \\ \therefore \textrm{ }\text{T} = \left(273\right) \times \left(5\right)^{\frac{2}{5}} \\ < \left(K E\right)_{rotatonal} > = \text{kT} \\ = 1.38 \times 10^{- 23} \times 273 \times \left(5\right)^{\frac{2}{5}} \\ = 7.158 \times 10^{- 21}   \text{J}\textrm{ } = 7 \times 10^{- 21}   \text{J} \left(\text{approx}\right) .\end{matrix}$

Correct Answer: 3600

$k_{i} + u_{i} = k_{f} + u_{f}$ $\frac{1}{2} m_{gas} v^{2} + \frac{f}{2} nRT_{i} = 0 + \frac{f}{2} nRT_{f}$ $\frac{3}{2} nR \left(T_{f} - T_{i}\right) = \frac{1}{2} m_{gas} v^{2}$ $\frac{3}{2} \left(1\right) R \left(ΔT\right) = \frac{1}{2} \left(4\right) \left(30\right)^{2}$ $ΔT = \frac{1200}{R} = \frac{x}{3 R} \Rightarrow x = 3600$

Correct Answer: 370

$ΔU = \left(\frac{f_{1}}{2}\right) \left(\eta_{1}\right) \left(R\right) \left(T\right) + \left(\frac{f_{2}}{2}\right) \left(x_{2}\right) \left(R\right) \left(T\right)$ $ΔU = \left(\frac{5}{2}\right) \left(5\right) \left(R\right) \left(20\right) + \left(\frac{3}{2}\right) \left(4\right) \left(R\right) \left(20\right)$ $ΔU = 250 R + 120 R$ $ΔU = 370 R = xR$ $x = 370$

Correct Option: (B)

$12 . 6   \times   10^{– 21} J ,   869   m / s$ $Average   translational   kinetic   energy   of   a   gas   molecule ,   KE_{av}   = \frac{3}{2} kT     or   KE_{av}   \propto   T$ $\therefore \frac{KE_{av}^{'}}{KE_{av}^{}} = \frac{T_{2}}{T_{1}} = \frac{1200}{400}$ $KE_{av}^{'} = 3 KE_{av}     =   3   \times   4 . 2   \times   10^{- 21}$ $=   12 . 6   \times   10^{- 21}   J$ $Also   v_{rrns}   = \sqrt{\frac{3 RT}{M}} or   v_{rrns}   \propto \sqrt{T}$ $\therefore \frac{v_{rms}^{'}}{v_{rms}} = \sqrt{\frac{1200}{400}}$ $\Rightarrow v_{rms}^{'} = \sqrt{3} v_{rms}$ $=   502 \sqrt{3}$ $=   869 . 464   m / s  $

Correct Option: (B)

Is proportional to the absolute temperature of the gas $E = \frac{3}{2} k T E T$

Correct Answer: 0.016

0.016 $\begin{matrix}T h e a v e r a g e t r a n s l a t i o n a l k i n e t i c e n e r g y p e r m o l c u l e i s \frac{3}{2} k T . \\ W h i c h d e p e n d o n t e m p e r a t u r e o n l y . T h e r e f o r t r a n s l a t i o n a l k i n e t i c \\ E n e r g y o f N_{2} a n d H_{2} b o t h w e l l b e e q u a l\end{matrix}$

Correct Answer: 81 J

81 J