A container of $10 \mathrm{~L}$ is filled with an ideal gas at a temperature of $27^{\circ} \mathrm{C}$ at a…

A container of $10 \mathrm{~L}$ is filled with an ideal gas at a temperature of $27^{\circ} \mathrm{C}$ at a pressure $12 \mathrm{~atm}$. The volume of the container is reduced to $6 \mathrm{~L}$ and the temperature of the gas in increased by $30^{\circ} \mathrm{C}$, then the final pressure of the gas is
  1. $22 \mathrm{~atm}$
  2. $20 \mathrm{~atm}$
  3. $11 \mathrm{~atm}$
  4. $9 \mathrm{~atm}$

Solution

Given, $V_1=10 \mathrm{~L}=10 \times 10^{-3} \mathrm{~m}^3$ $ \begin{aligned} & \mathrm{T}_1=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K} \\ & p_1=12 \mathrm{~atm} \\ & V_2=6 \mathrm{~L}=6 \times 10^{-3} \mathrm{~m}^3 \\ & T_2=30^{\circ} \mathrm{C}=273+30=303 \mathrm{~K} \\ & p_2=? \end{aligned} $ By ideal gas equation, $ \begin{aligned} \frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} \\ \Rightarrow \quad p_2 & =\frac{p_1 V_1 T_2}{T_1 V_2} \\ & =\frac{12 \times 10 \times 10^{-3} \times 303}{300 \times 6 \times 10^{-3}}=20.2 \mathrm{~atm} \\ & \simeq 20 \mathrm{~atm} \end{aligned} $

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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