A monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ at a pressure of $4 \mathrm{~atm}$ is compressed…
A monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ at a pressure of $4 \mathrm{~atm}$ is compressed adiabatically so that its temperature rises from $27^{\circ} \mathrm{C}$ to $327^{\circ} \mathrm{C}$. The pressure of the gas in its final state is
$2^{\frac{5}{3}} \mathrm{~atm}$
$2^{\frac{10}{3}} \mathrm{~atm}$
$2^{\frac{5}{2}}$ atm
$2^{\frac{9}{2}} \mathrm{~atm}$
Solution
(1) For an adiabatic change,
$
\begin{aligned}
& p V^\gamma=\text { constant } \\
\Rightarrow \quad & T^\gamma p^{1-\gamma}=\text { constant or } T_1^\gamma, p_1^{1-\gamma}=T_2^\gamma \cdot p_2^{1-\gamma} \\
\Rightarrow \quad & p_2^{1-\gamma}=\left(\frac{T_1}{T_2}\right)^\gamma \cdot p_1^{1-\gamma} \\
\Rightarrow \quad & p_2=\left(\frac{T_1}{T_2}\right)^{\frac{\gamma}{1-\gamma}} \cdot p_1
\end{aligned}
$
Here,
$
\begin{aligned}
& p_1=4 \mathrm{~atm} ; \gamma=5 / 3 \\
& T_1=27^{\circ} \mathrm{C}=300 \mathrm{~K} \\
& T_2=327^{\circ} \mathrm{C}=600 \mathrm{~K}
\end{aligned}
$
Now, substituting values in Eq. (i), we get
$
\begin{aligned}
p_2 & =\left(\frac{300}{600}\right)^{\frac{5 / 3}{1-5 / 3}} \times 4=\left(\frac{1}{2}\right)^{\frac{-5}{2}} \times 2^2 \\
& =2^{2+\frac{5}{2}}=2^{\frac{9}{2}}
\end{aligned}
$
Hence, the pressure of the gas in its final state is $2^{9 / 2} \mathrm{~atm}$