Question 123: 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 Ti (in Kelvin) and the final temperature is a Ti, 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 ${1 \over {32}} of its initial volume. If the initial temperature of the gas is Ti (in Kelvin) and the final temperature is a Ti, the value of a$ is

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