Q134 / 180 · COMEDK 2020 · Physics
$1 \mathrm{~mm}^{3}$ of a gas is compressed at 1 atmospheric pressure and temperature $27^{\circ} \mathrm{C}$ to $627^{\circ} \mathrm{C}$. What is the final pressure under adiabatic condition? $(\gamma$ for the gas $=15$ )
- Correct answerA$27 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$
- B$80 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$
- C$36 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$
- D$56 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$
Solution
Given, $V_{1}=1 \mathrm{~mm}^{3}$
$$
\begin{aligned}
&=10^{-9} \mathrm{~m}^{3} \\
p_{1} &=1 \mathrm{~atm} \\
&=1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} \\
T_{1} &=27+273 \\
&=300 \mathrm{~K} \\
T_{2} &=627+273 \\
&=900 \mathrm{~K}
\end{aligned}
$$
Under adiabatic condition,
$$
\begin{aligned}
& p_{1}^{1-\gamma} T_{1}^{\gamma}=p_{2}^{1-\gamma} T_{2}^{\gamma} \\
\Rightarrow \quad\left(\frac{p_{2}}{p_{1}}\right)^{1-\gamma} &=\left(\frac{T_{1}}{T_{2}}\right)^{\gamma} \\
\Rightarrow \quad\left(\frac{p_{2}}{p_{1}}\right)^{1-1.5} &=\left(\frac{300}{900}\right)^{1.5} \\
\Rightarrow \quad &\left(\frac{p_{2}}{p_{1}}\right)^{-0.5}=\left(\frac{1}{3}\right)^{1.5} \\
\Rightarrow \quad &\left(\frac{p_{1}}{p_{2}}\right)^{0.5}=\left(\frac{1}{3}\right)^{1.5}
\end{aligned}
$$
$\Rightarrow \frac{p_{1}}{p_{2}}=\left(\frac{1}{3}\right)^{\frac{15}{05}}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$
$\Rightarrow \quad p_{2}=27 p_{1}=27 \times 1 \times 10^{5}$
$=27 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$
Hence, the correct answer is option: 'A'.