Question

5.0 moles of an Ideal gas at 3.0 atm pressure and 27$^\circ$C is compressed isothermally to half its volume by application of an external pressure of 3.5 atm. What is the amount of work done (in joules) on the gas? Given: 1 L atm = 101.3 J, R = 0.082 L atm K$^{-1}$ mol$^{-1}$

• A. -3559.9 J
• B. 7268.3 J
• C. -10367.4 J
• D. 14359.2 J

Hint:

In isothermal processes, temperature remains constant, and the work done can be directly related to the change in volume. Be sure to use the correct units and conversion factors.

Answer 1

The Correct Option is B

For an isothermal process, the work done on the gas can be calculated using the formula: \[ W = - P_{\text{ext}} \Delta V \] Where: - \( P_{\text{ext}} \) is the external pressure (3.5 atm), - \( \Delta V \) is the change in volume. For an ideal gas undergoing isothermal compression, the change in volume is related to the initial and final pressures and volumes. The volume of the gas is reduced by half, so: \[ V_f = \frac{V_i}{2} \] From the ideal gas law \( PV = nRT \), we know: \[ V_i = \frac{nRT}{P_i} \] Where: - \( n = 5.0 \, \text{mol} \), - \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \), - \( T = 27^\circ C = 300 \, \text{K} \), - \( P_i = 3.0 \, \text{atm} \). First, calculate the initial volume \( V_i \): \[ V_i = \frac{(5.0)(0.082)(300)}{3.0} = 409.0 \, \text{L} \] Now, the final volume \( V_f \) will be: \[ V_f = \frac{V_i}{2} = \frac{409.0}{2} = 204.5 \, \text{L} \] Now, calculate the work done: \[ \Delta V = V_f - V_i = 204.5 - 409.0 = -204.5 \, \text{L} \] \[ W = - P_{\text{ext}} \Delta V = - (3.5 \, \text{atm}) (-204.5 \, \text{L}) = 7268.3 \, \text{L atm} \] To convert this to joules, multiply by 101.3 J per L atm: \[ W = 7268.3 \, \text{L atm} \times 101.3 \, \text{J/L atm} = 7268.3 \, \text{J} \]
Thus, the amount of work done is 7268.3 J.