Question

An ideal gas expands against a constant external pressure of 2.0 atmosphere from 20 litre to 40 litre and absorbs 10 kJ of heat from the surrounding. What is the change in internal energy of the system? (given: \( 1 \, {atm-litre} = 101.3 \, {J} \))

• A. 4052 J
• B. 5948 J
• C. 14052 J
• D. 9940 J

Hint:

The work done during expansion at constant pressure is calculated as \( W = P_{{ext}} \times \Delta V \), and the change in internal energy is \( \Delta U = Q - W \).

Answer 1

The Correct Option is B

Step 1: First, calculate the work done by the gas during the expansion. The work done by the gas during expansion at constant pressure is given by: \[ W = P_{{ext}} \times \Delta V. \] Substitute the given values: \[ W = 2.0 \, {atm} \times (40 \, {L} - 20 \, {L}) = 2.0 \, {atm} \times 20 \, {L} = 40 \, {atm-L}. \] Step 2: Convert the work to Joules using the conversion factor \( 1 \, {atm-L} = 101.3 \, {J} \): \[ W = 40 \, {atm-L} \times 101.3 \, {J/atm-L} = 4052 \, {J}. \] Step 3: The first law of thermodynamics states: \[ \Delta U = Q - W, \] where \( \Delta U \) is the change in internal energy, \( Q \) is the heat absorbed, and \( W \) is the work done. Substitute the values: \[ \Delta U = 10000 \, {J} - 4052 \, {J} = 5948 \, {J}. \]