Question:
A container of \(10\,\text{L}\) is filled with an ideal gas at a temperature of \(27^\circ\text{C}\) at a pressure of \(12\,\text{atm}\). The volume of the container is reduced to \(6\,\text{L}\) and the temperature of the gas is increased by \(30^\circ\text{C}\), then the final pressure of the gas is
A container of \(10\,\text{L}\) is filled with an ideal gas at a temperature of \(27^\circ\text{C}\) at a pressure of \(12\,\text{atm}\). The volume of the container is reduced to \(6\,\text{L}\) and the temperature of the gas is increased by \(30^\circ\text{C}\), then the final pressure of the gas is
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In gas law problems, always convert temperature from Celsius to Kelvin before substitution:
\[
T(K)=T(^\circ C)+273
\]
Updated On: Jun 26, 2026
- \(22\,\text{atm}\)
- \(20\,\text{atm}\)
- \(11\,\text{atm}\)
- \(9\,\text{atm}\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the combined gas law.
For a fixed amount of ideal gas, \[ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \]
Step 2: Write the given values.
Initial pressure: \[ P_1=12\,\text{atm} \] Initial volume: \[ V_1=10\,\text{L} \] Initial temperature: \[ T_1=27^\circ\text{C}=300\,\text{K} \] Final volume: \[ V_2=6\,\text{L} \] Temperature is increased by \(30^\circ\text{C}\), so \[ T_2=57^\circ\text{C}=330\,\text{K} \]
Step 3: Substitute in the formula.
\[ \frac{12\times 10}{300}=\frac{P_2\times 6}{330} \] So, \[ P_2=\frac{12\times 10\times 330}{300\times 6} \] \[ P_2=22\,\text{atm} \]
Step 4: Final conclusion.
Hence, the final pressure of the gas is \[ \boxed{22\,\text{atm}} \]
For a fixed amount of ideal gas, \[ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \]
Step 2: Write the given values.
Initial pressure: \[ P_1=12\,\text{atm} \] Initial volume: \[ V_1=10\,\text{L} \] Initial temperature: \[ T_1=27^\circ\text{C}=300\,\text{K} \] Final volume: \[ V_2=6\,\text{L} \] Temperature is increased by \(30^\circ\text{C}\), so \[ T_2=57^\circ\text{C}=330\,\text{K} \]
Step 3: Substitute in the formula.
\[ \frac{12\times 10}{300}=\frac{P_2\times 6}{330} \] So, \[ P_2=\frac{12\times 10\times 330}{300\times 6} \] \[ P_2=22\,\text{atm} \]
Step 4: Final conclusion.
Hence, the final pressure of the gas is \[ \boxed{22\,\text{atm}} \]
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