Question 287: Thermodynamics & KTG

Question

A sample of gas at temperature $$T$$ is adiabatically expanded to double its volume. The work done by the gas in the process is $$\left(\mathrm{given}, \gamma=\frac{3}{2}\right)$$ :

Accepted Answer

$\gamma=\frac{3}{2}$
$$
\begin{aligned}
& W =\frac{n R \Delta T}{1-\gamma}=\frac{n R T_{f}-n R T_{i}}{1-\gamma} \\
& =\frac{(P V)_{f}-\left(P V_{i}ight)}{1-\gamma} \quad \ldots 	ext { (1) } \\
& P V^{\gamma}=	ext { constant } \\
& P_{i} V_{i}^{\gamma}=P_{f}\left(2 V_{i}ight)^{\gamma} \Rightarrow P_{f}=\frac{P_{i}}{2^{\gamma}}=\frac{P_{i}}{2 \sqrt{2}} ......(2)
\end{aligned}
$$
From (1) and (2)
$$
\begin{aligned}
& W=\frac{\frac{P_{i}}{2 \sqrt{2}} 2 V_{i}-P_{i} V_{i}}{1-\gamma}=\frac{P_{i} V_{i}}{-1 / 2}\left(\frac{1}{\sqrt{2}}-1ight) \\
& =-n R T(\sqrt{2}-2) \\
& =n R T(2-\sqrt{2})
\end{aligned}
$$

Answer

$$W=T R[\sqrt{2}-2]$$

Answer

$$W=\frac{T}{R}[\sqrt{2}-2]$$

Answer

$$W=\frac{R}{T}[2-\sqrt{2}]$$

Answer

$$W=R T[2-\sqrt{2}]$$

Thermodynamics & KTG

A sample of gas at temperature $T is adiabatically expanded to double its volume. The work done by the gas in the process is \left(\mathrm{given}, \gamma=\frac{3}{2}\right)$ :

Choose the correct answer: