Question

The volume of one mole of an ideal gas with the abiabatic exponent $\gamma$ is varied according to the law $V=\frac{a}{T}$, where $a$ is a gas constant. Find the amount of heat obtained by the gas in this process, if the gas temperature is increased by $\Delta T$. $R$ is a gas constant

• A. $nR\Delta T\left(\frac{2-\gamma }{\gamma -1}\right]$
• B. $nR\Delta T\left(\frac{2+\gamma }{\gamma -1}\right]$
• C. $nR\Delta T\left(\frac{2-\gamma }{\gamma +1}\right]$
• D. $nR\Delta T\left(\frac{2+\gamma }{\gamma +1}\right]$

Answer 1

The correct option is A $nR\Delta T\left(\frac{2-\gamma }{\gamma -1}\right]$

For an ideal gas $PV=nRT$
$P=\frac{nRT}{V}=\frac{nR{T}^2}{a}dV=-\frac{a}{T^2}dT$
Work done
$dW=PdV=\frac{nR{T}^2}{a}\left(\frac{-a}{T^2}dT\right)=-nRdT$
Integrating above equation, we get
$W=\int dW={\int }_T^{T+\Delta T}nR\Delta T=-nR\Delta T$
We define change in internal energy
$\Delta U=n\left(\frac{R}{\gamma -1}\right)\Delta T$
From first law of thermodynamics
$Q=\Delta U+W=\frac{nR\Delta T}{\gamma -1}-nR\Delta T$
$Q=nR\Delta T\left(\frac{1}{\gamma -1}-1\right]=nR\Delta T\left(\frac{2-\gamma }{\gamma -1}\right]$