Q139 / 160 · TS EAMCET 2019 (03 May Shift 1) · Physics
One mole of nitrogen gas being initially at a temperature of $T_0=300 \mathrm{~K}$ is adiabatically compressed to increase its pressure 10 times. The final gas temperature after compression is (Assume, nitrogen gas molecules as rigid diatomic and $100^{1 / 7}=1.9$ )
- A$120 \mathrm{~K}$
- B$750 \mathrm{~K}$
- C$650 \mathrm{~K}$
- Correct answerD$570 \mathrm{~K}$
Solution
Given, initial temperature of 1 mole of $\mathrm{N}_2$ gas
$$
r_0=300 \mathrm{~K}
$$
initial pressure of gas, $p_1=p$
and final pressure of gas, $p_2=10 p$
By adiabatic process relation between $p$ and $T$,
$$
\begin{aligned}
p_1^{1-\gamma} T_1^\gamma & =p_2^{1-\gamma} \cdot \mathrm{T}_2^\gamma \\
\left(\frac{p_1}{p_2}\right)^{1-\gamma} & =\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^\gamma \Rightarrow\left(\frac{p}{10 p}\right)^{1-\gamma}=\left(\frac{\mathrm{T}_2}{300}\right)^\gamma \\
10^{\gamma-1} & =\frac{\mathrm{T}_2^\gamma}{300^\gamma} \Rightarrow \mathrm{T}_2^\gamma=10^{\gamma-1} \times 300^\gamma \\
\Rightarrow \quad \mathrm{T}_2 & =10^{\frac{\gamma-1}{\gamma}} \times 300=10^{1-\frac{1}{\gamma}} \times 300 \\
& =10^{1-\frac{1}{7 / 5}} \times 300\left[\because \text { For diatomic, } \gamma=\frac{7}{5}\right] \\
& =10^{2 / 7} \times 300=(100)^{1 / 7} \times 300 \\
& =1.9 \times 300 \\
& =570 \mathrm{~K}
\end{aligned}
$$
Therefore, the final gas temperature after compression is $570 \mathrm{~K}$.
Hence, the correct answer is option: 'D'.