Q80 / 120 · NEET 2011 (Mains) · Physics
A mass of diatomic gas $(\gamma=1.4)$ at a pressure of $2 \mathrm{~atm}$ is compressed adiabatically so that its temperature rise from $27^{\circ} \mathrm{C}$ to $927^{\circ} \mathrm{C}$. The pressure of the gas is final state is
- A$28 \mathrm{~atm}$
- B$68.7 \mathrm{~atm}$
- Correct answerC$256 \mathrm{~atm}$
- D$8 \mathrm{~atm}$
Solution
$\begin{aligned}
& T_1=273+27=300 \mathrm{~K} \\
& T_2=273+927=1200 \mathrm{~K}
\end{aligned}$
Gas equation for adiabatic process
$\begin{aligned}
& p V^\gamma=\text { constant } \\
& p\left(\frac{T}{p}\right)^\gamma=\text { constant } \quad(\because p V=R T) \\
& \therefore \quad \frac{p_2}{p_1}=\left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}} \\
& p_2=p_1\left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}} \\
& p_2=2\left(\frac{1200}{300}\right)^{\frac{1.4}{1.4-1}} \\
& p_2=256 \mathrm{~atm}
\end{aligned}$
Hence, the correct answer is option: 'C'.