Question 1: Thermodynamics & KTG

Question

A diatomic ideal gas is compressed adiabatically ${1 \over {32}}$ of its initial volume. If the initial temperature of the gas is T
i
(in Kelvin) and the final temperature is a T
i
, the value of $a$ is

Accepted Answer

The diatomic gas has five degrees of freedom i.e., f = 5. Thus, the internal energy per mole, specific heat at constant volume, specific heat at constant pressure, and the ratio of specific heats for a diatomic gas, are given by
$U = (f/2)RT = (5/2)RT$, ${C_v} = dU/dT = 5R/2$, ${C_p} = {C_v} + R = 7R/2$, $\gamma  = {C_p}/{C_v} = 7/5$.
For an adiabatic process, $T{V^{\gamma  - 1}}$ = constant
${T_i}V_i^{\gamma  - 1} = {T_f}V_f^{\gamma  - 1}$
Substituting the given values, we get
${T_i}V_i^{\gamma  - 1} = a{T_i}{\left( {{{{V_i}} \over {32}}} \right)^{\gamma  - 1}} \Rightarrow a = {32^{\gamma  - 1}}$
For diatomic gas, $\gamma  = {7 \over 5}$
$a = {32^{{7 \over 5} - 1}} = {32^{2/5}} = {2^2} = 4$

Thermodynamics & KTG

A diatomic ideal gas is compressed adiabatically {1 \over {32}} of its initial volume. If the initial temperature of the gas is T i (in Kelvin) and the final temperature is a T i , the value of a is

Integer / Numeric Answer Type