Q. An ideal gas at $27 \, ^\circ C$ is compressed adiabatically to $8/27$ of its original volume. If $\gamma =5/3$ , then the rise in temperature is
NTA AbhyasNTA Abhyas 2022
Solution:
For an adiabatic process $TV^{\gamma - 1}=\text{constant}$ ; where $T$ is temperature and $V$ denotes the volume, therefore,
$\frac{T_{1}}{T_{2}}=\left[\frac{V_{2}}{V_{1}}\right]^{\gamma - 1}$
$T_{2}=T_{1}\left[\frac{V_{1}}{V_{2}}\right]^{\gamma - 1}$
$T_{2}=300\left[\frac{2 7}{8}\right]^{5 / 3 - 1}=300\left[\frac{2 7}{8}\right]^{2 / 3}=675K$
Hence, the change in temperature,
$\Delta T=675-300=375K$ .
$\frac{T_{1}}{T_{2}}=\left[\frac{V_{2}}{V_{1}}\right]^{\gamma - 1}$
$T_{2}=T_{1}\left[\frac{V_{1}}{V_{2}}\right]^{\gamma - 1}$
$T_{2}=300\left[\frac{2 7}{8}\right]^{5 / 3 - 1}=300\left[\frac{2 7}{8}\right]^{2 / 3}=675K$
Hence, the change in temperature,
$\Delta T=675-300=375K$ .