Question:medium

A gas occupies a volume of 10.0 L at a pressure of 2.0 atm and a temperature of 300 K. What will the volume be if the pressure is increased to 4.0 atm and the temperature is increased to 600 K? (Assume the amount of gas remains constant.)

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The combined gas law allows you to solve for any of the gas properties (pressure, volume, or temperature) when the others change. Always ensure the units are consistent, and use the correct relationships.
Updated On: Nov 26, 2025
  • \( 5.0 \, \text{L} \)
  • \( 10.0 \, \text{L} \)
  • \( 20.0 \, \text{L} \)
  • \( 2.5 \, \text{L} \)
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The Correct Option is A

Solution and Explanation

Step 1: Apply the combined gas law. The combined gas law, derived from the ideal gas law, establishes a relationship between pressure, volume, and temperature: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] where: - \( P_1 \), \( V_1 \), and \( T_1 \) represent the initial pressure, volume, and temperature, respectively. - \( P_2 \), \( V_2 \), and \( T_2 \) represent the final pressure, volume, and temperature, respectively. Step 2: Input the known values. Provided data: - \( P_1 = 2.0 \, \text{atm} \) - \( V_1 = 10.0 \, \text{L} \) - \( T_1 = 300 \, \text{K} \) - \( P_2 = 4.0 \, \text{atm} \) - \( T_2 = 600 \, \text{K} \) The objective is to determine \( V_2 \). \[ \frac{(2.0)(10.0)}{300} = \frac{(4.0)(V_2)}{600} \] Step 3: Calculate \( V_2 \). \[ \frac{20.0}{300} = \frac{4.0 V_2}{600} \] \[ \frac{1}{15} = \frac{2 V_2}{300} \] \[ V_2 = \frac{1}{15} \times 150 = 5.0 \, \text{L} \] Answer: The final volume of the gas following the changes is \( 5.0 \, \text{L} \).
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