Question

An ideal gas of adiabatic exponent $\gamma$ is expanded so that the amount of heat transferred to the gas is equal to decrease in its internal energy. Then the equation of the process in terms of the variables temperature $T$ and volume $V$ is

• A. $T{V}^{\frac{(\gamma -1)}{2}}=C$
• B. $T{V}^{\frac{(\gamma -2)}{2}}=C$
• C. $T{V}^{\frac{(\gamma -1)}{4}}=C$
• D. $T{V}^{\frac{(\gamma -2)}{4}}=C$

Answer 1

The correct option is A $T{V}^{\frac{(\gamma -1)}{2}}=C$

Given: $dQ=-dU$
From first law of Thermodynamics
$dQ=dU+dW\dots \left(1\right)$
From equation (1) and (2)
$-dU=dU+dW$
$2dU+dW=0$
$\therefore 2\left[n{C}_{v}dt\right]+PdV=0$ $\{\because dU=n{C}_{v}dT\}$
$2\left[n\left(\frac{R}{\gamma -1}\right)dT\right]+PdV=0$
$\frac{2nRdT}{\gamma -1}+\left(\frac{nRT}{V}\right)dV=0$
or $\left(\frac{2}{\gamma -1}\right)\frac{dT}{T}+\frac{dV}{V}=0$
Integrating we get,
$\left(\frac{2}{\gamma -1}\right)ln\left(T\right)+ln\left(V\right)=ln\left(C\right)$
Solving, we get
$T{V}^{\frac{\gamma -1}{2}}=\text{constant}\left(C\right)$