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Q. If an ideal gas at $ 27 ^{\circ}C $ is compressed suddenly to one fourth of its initial volume, then rise in its temperature is $ (\gamma = 7/5) $

Thermodynamics

Solution:

Here, $ T_1 = 27^{\circ}C= 27 +273 = 300\,K $
$ V_2 = \frac{V_1}{4} $
As the gas is compressed suddenly, then the process is adiabatic, then
$ TV^{\gamma-1} = $ constant
$ \Rightarrow T_1 V_1 ^{\gamma-1} = T_2 V_2 ^{\gamma-1} $
or $ T_2 =T_1 (\frac{V_1}{V_2}) ^{\gamma-1} $
$ = 300 (\frac{V_1} {V_1/4})^{\frac{7}{5} -1} $
$ = 300 \times (4)^{2/5} $
$ = 522.33\,K $
Hence, Change in temperature
$ \Rightarrow \Delta T =T_2 -T_1 = 522.33 -300 $
$ = 222.33\,K $