Question

An ideal gas underwent a process in which it absorbed 7.5 J of energy as heat and undergoes an expansion in which it performed 5.0 J of work on its surroundings. If the expansion was performed against a pressure of 101325 Pa, what was the volume change (ΔV) for the expansion? Express your result in units of m^3. (Note: 1 J = 1 Pa m^3).

Answer 1

Nothing is mentioned about the temperature change of the gas; hence, I assume the process is isothermal.

The work is done on the surroundings; hence, work is done by the system and is therefore, negative; consequently, we have w = -5.0 J.

Since only pressure-volume work is done, we have w = -P_ext*ΔV where P_ext = 101325 Pa. Therefore,

-5.0 J = -(101325 Pa)*ΔV

====> (5.0 J)*(1 Pa.m³/1 J) = (101325 Pa)*ΔV (1 Pa = 1 J.m³)

====> ΔV = 4.9346*10^-5 m³ ≈ 4.935*10^-5 m³

The expansion in volume of the gas is 4.935*10^-5 m³ (ans).