Q152 / 155 · WBJEE 2019 · Physics
The initial pressure and volume of a given mass of an ideal gas $\left(\right.$ with $\frac{C_{p}}{C_{V}}=\gamma),$ taken in a cylinder fitted with a piston, are $p_{0}$ and $V_{0}$ respectively. At this stage the gas has the same temperature as that of the surrounding medium which is $T_{0}$. It is adiabatically compressed to a volume equal to $\frac{V_{0}}{2}$ Subsequently the gas is allowed to come to thermal equilibrium with the surroundings. What is the heat released to the surrounding?
- A0
- Correct answerB$(2^{\gamma-1}-1) \frac{p_{0} V_{0}}{\gamma-1}$
- C$\gamma {p_{0}} {V_{0}} ln2$
- D$\frac{{p_{0}} V_{0}}{2(\gamma-1)}$
Solution
The equation of state for an ideal gas undergoing adiabatic process is given as,
$$
T V^{\gamma-1}=\text { constant }
$$
Let the temperature after adiabatic compression as given in the question be $T$ then,
$$
T_{0} V^{\gamma-1}=T\left(\frac{V_{0}}{2}\right)^{\gamma-1}
$$
$\Rightarrow \quad T=T_{0} 2^{\gamma-1}$
Now, heat released at volume $\frac{V_{0}}{2}$ to achieve
temperature $T_{0}$ The net heat released can be determined by the
equation.
$$
\begin{aligned}
\Delta Q &=\mu C_{V} \Delta T \\
&=\mu \times \frac{R}{\gamma-1} \quad\left(T_{0} 2^{\gamma-1}-T_{0}\right) \\
&=\frac{\mu R T_{0}}{\gamma-1}\left(2^{\gamma-1}-1\right)\left(\because \mu T_{0} R=p_{0} V_{0}\right)
\end{aligned}
$$
$\therefore$ Heat released $=\frac{p_{0} V_{0}}{\gamma-1}\left(2^{\gamma-1}-1\right)$
Hence, the correct answer is option: 'B'.