MHT CET · Physics · Thermodynamics
A gas is contained in closed vessel. The initial temperature of the gas is \(100^{\circ} \mathrm{C}\). If the pressure of the gas is increased by \(4 \%\), the increase in the temperature of the gas is
- A \(2^{\circ} \%\)
- B \(3^{\circ} \%\)
- C \(4^{\circ} \%\)
- D \(5^{\circ} \%\)
Answer & Solution
Correct Answer
(C) \(4^{\circ} \%\)
Step-by-step Solution
Detailed explanation
Given:
- Initial temperature \(T_1=100^{\circ} \mathrm{C}=373 \mathrm{~K}\),
- Pressure increases by \(4 \%: P_2=P_1+0.04 P_1=1.04 P_1\).
Formula: From the ideal gas equation \(P V=n R T\), at constant volume:
\(\begin{gathered}
\frac{T_2}{T_1}=\frac{P_2}{P_1} \\
T_2=T_1 \cdot \frac{P_2}{P_1}
\end{gathered}\)
Step 1: Substitute the values:
\(\begin{gathered}
T_2=373 \cdot \frac{1.04 P_1}{P_1}=373 \cdot 1.04 \\
T_2=387.92 \mathrm{~K}
\end{gathered}\)
Step 2: Increase in Temperature:
\(\Delta T=T_2-T_1=387.92-373=14.92 \mathrm{~K} \approx 15^{\circ} C\)
Answer: \(4\% C\), Option 3.
- Initial temperature \(T_1=100^{\circ} \mathrm{C}=373 \mathrm{~K}\),
- Pressure increases by \(4 \%: P_2=P_1+0.04 P_1=1.04 P_1\).
Formula: From the ideal gas equation \(P V=n R T\), at constant volume:
\(\begin{gathered}
\frac{T_2}{T_1}=\frac{P_2}{P_1} \\
T_2=T_1 \cdot \frac{P_2}{P_1}
\end{gathered}\)
Step 1: Substitute the values:
\(\begin{gathered}
T_2=373 \cdot \frac{1.04 P_1}{P_1}=373 \cdot 1.04 \\
T_2=387.92 \mathrm{~K}
\end{gathered}\)
Step 2: Increase in Temperature:
\(\Delta T=T_2-T_1=387.92-373=14.92 \mathrm{~K} \approx 15^{\circ} C\)
Answer: \(4\% C\), Option 3.
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