A gas is contained in closed vessel. The initial temperature of the gas is $100^{\circ} \mathrm{C}$. If the…

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
  1. $2^{\circ} \%$
  2. $3^{\circ} \%$
  3. $4^{\circ} \%$
  4. $5^{\circ} \%$

Solution

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.

Asked in: MHT CET 2024 (09 May Shift 1)

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