Question

An ideal gas is contained in a cylinder with a movable piston. The gas undergoes an isothermal expansion from a volume of 2 liters to 8 liters at a temperature of 300 K. If the initial pressure of the gas is \( 2 \times 10^5 \, \text{Pa} \), calculate the work done by the gas during this expansion. (Use \( R = 8.31 \, \text{J/mol·K} \))

• A. \( 7200 \, \text{J} \)
• B. \( 3600 \, \text{J} \)
• C. \( 1800 \, \text{J} \)
• D. \( 14400 \, \text{J} \)

Hint:

For isothermal processes, the work done by the gas can be calculated using the equation \( W = nRT \ln \left( \frac{V_f}{V_i} \right) \). Always remember to convert units where necessary (e.g., liters to cubic meters).

Answer 1

The Correct Option is A

For an isothermal expansion of an ideal gas, work done is \( W = nRT \ln \left( \frac{V_f}{V_i} \right) \). Parameters: - \( W \): work done by the gas - \( n \): number of moles of the gas - \( R = 8.31 \, \text{J/mol·K} \): universal gas constant - \( T = 300 \, \text{K} \): temperature - \( V_f = 8 \, \text{L} \): final volume - \( V_i = 2 \, \text{L} \): initial volume First, determine the number of moles \( n \) using the ideal gas law: \( PV = nRT \). Substituting given values: \( (2 \times 10^5) \times (2 \times 10^{-3}) = n \times 8.31 \times 300 \). Thus, \( n = \frac{(2 \times 10^5) \times (2 \times 10^{-3})}{8.31 \times 300} \approx 0.16 \, \text{mol} \). Next, calculate the work done: \( W = 0.16 \times 8.31 \times 300 \times \ln \left( \frac{8}{2} \right) \) \( W = 0.16 \times 8.31 \times 300 \times \ln(4) \) \( W \approx 7200 \, \text{J} \). The work done by the gas is \( 7200 \, \text{J} \).