# Kinetic Theory - Class 11 - Physics

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## Extra Questions - Kinetic Theory | NCERT | Physics | Class 11

For an ideal gas, the slope of the $V$-$T$ graph during an adiabatic process is $\frac{dV}{dT}=-m$ at a point where volume and temperature are $V_{0}$ and $T_{0}$. Find the value of $C_{p}$ of the gas. It is given that $\mathrm{m}$ is a positive number.

A. $\left(\frac{m T_{0}}{V_{0}}+1\right) R$

B. $\frac{m T_{0} R}{V_{0}}$

C. $\left(\frac{-m T_{0}}{V_{0}}+1\right) R$

D. $\left(\frac{-m T_{0} R}{V_{0}}\right)$

The correct option is **A.** $\left(\frac{m T_{0}}{V_{0}}+1\right) R$

For an adiabatic process, the equation that relates temperature $T$ and volume $V$ can be represented as: $$ TV^{\gamma-1} = \text{constant} $$ Taking the natural logarithm on both sides gives: $$ \ln T + (\gamma - 1) \ln V = \text{constant} $$ Differentiating both sides with respect to $T$, we obtain: $$ \frac{1}{T} + (\gamma - 1) \frac{dV}{V} = 0 $$ Rearranging, we have: $$ \frac{dV}{dT} = -\frac{V}{T(\gamma - 1)} $$

Given that $\left(\frac{dV}{dT}\right){V{0}, T_{0}} = -m$, equating it to the expression from the derivative we get: $$ -\frac{V_0}{T_0(\gamma - 1)} = -m $$ This implies: $$ \frac{1}{\gamma - 1} = \frac{m T_{0}}{V_{0}} $$

We know the relation $C_v = \frac{R}{\gamma-1}$. Substituting what we have found:
$$
C_v = \frac{m R T_{0}}{V_{0}}
$$
And the heat capacity at constant pressure, $C_p$, is given by $C_v + R$:
$$
C_p = \left(\frac{m T_{0}}{V_{0}} + 1\right) R
$$
Thus, the answer is **option A.**

Who among the following proposed the Big Bang Theory?

A. Albert Einstein

B. Georges Lemaitre

C. Charles Darwin

D. Francesco Redi

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If the number of molecules of $\mathrm{H}*{2}$ are double that of $\mathrm{O}*{2}$, then the ratio of kinetic energy of hydrogen to that of oxygen at $300 \mathrm{~K}$ is

(A) $1: 1$

(B) $1: 2$

(C) $2: 1$

(D) $1: 16$