Question

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.)

• A. \( 5.0 \, \text{L} \)
• B. \( 10.0 \, \text{L} \)
• C. \( 20.0 \, \text{L} \)
• D. \( 2.5 \, \text{L} \)

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Use the combined gas law. The combined gas law is derived from the ideal gas law and relates pressure, volume, and temperature: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] where: - \( P_1 \), \( V_1 \), and \( T_1 \) are the initial pressure, volume, and temperature, - \( P_2 \), \( V_2 \), and \( T_2 \) are the final pressure, volume, and temperature. Step 2: Substitute the known values. Given: - \( P_1 = 2.0 \, \text{atm} \), - \( V_1 = 10.0 \, \text{L} \), - \( T_1 = 300 \, \text{K} \), - \( P_2 = 4.0 \, \text{atm} \), - \( T_2 = 600 \, \text{K} \). We need to solve for \( V_2 \). \[ \frac{(2.0)(10.0)}{300} = \frac{(4.0)(V_2)}{600} \] Step 3: Solve for \( V_2 \). \[ \frac{20.0}{300} = \frac{4.0 V_2}{600} \] \[ \frac{1}{15} = \frac{2 V_2}{300} \] \[ V_2 = \frac{1}{15} \times 150 = 5.0 \, \text{L} \] Answer: Therefore, the volume of the gas after the changes is \( 5.0 \, \text{L} \).