Question

2 L of a gas at 1 atm pressure is reversibly heated to reach a final volume of 3.5 L. The absolute value of the work done on the gas (rounded off to the nearest integer) is \(\underline{\hspace{2cm}}\) Joules.

Hint:

The work done on a gas during an isothermal process can be calculated using the formula \( W = -nRT \ln \left( \frac{V_f}{V_i} \right) \).

Answer 1

Correct Answer: 151

The work done on the gas during an isothermal expansion is given by:
\[ W = -nRT \ln \left( \frac{V_f}{V_i} \right) \] where \( n \) is the number of moles, \( R \) is the gas constant, \( T \) is the temperature (constant), \( V_f \) is the final volume, and \( V_i \) is the initial volume. Assuming the temperature remains constant, we calculate:
\[ W = -1 \times 8.314 \times 298 \times \ln \left( \frac{3.5}{2} \right). \] Solving for \( W \), we get:
\[ W \approx -151 \, \text{J}. \] Thus, the work done on the gas is \( 151 \, \text{J} \).