Question

When 2 moles of an ideal gas are expanded isothermally from a volume of 12.5 L to 15.0 L against constant external pressure of 760 mm Hg, calculate the amount of work done in joule?

• A. $-253.25\ \text{J}$
• B. $-190.0\ \text{J}$
• C. $-1924.0\ \text{J}$
• D. $-25.325\ \text{J}$

Hint:

For constant external pressure, work depends only on pressure and volume change, not on the nature of the process.

Answer 1

The Correct Option is A

Step 1: Formula for work known external pressure.
For expansion against constant external pressure, work done is given by:
\[ W = -P_{\text{ext}} \Delta V \]
Step 2: Convert pressure into SI units.
Given external pressure = $760\ \text{mm Hg} = 1\ \text{atm}$
\[ 1\ \text{atm} = 101325\ \text{Pa} \]
Step 3: Calculate change in volume.
\[ \Delta V = V_2 - V_1 = 15.0 - 12.5 = 2.5\ \text{L} \] \[ 2.5\ \text{L} = 2.5 \times 10^{-3}\ \text{m}^3 \]
Step 4: Calculate work done.
\[ W = -(101325)(2.5 \times 10^{-3}) \] \[ W = -253.31\ \text{J} \approx -253.25\ \text{J} \]
Step 5: Conclusion.
The work done during isothermal expansion against constant external pressure is $-253.25\ \text{J}$.