A student is calculating the work done by 2 mole of an ideal gas in a reversible isothermal expansion shown in the figure. He by mistake calculated the area of the shaded area in the PV graph as work

Question

A student is calculating the work done by 2 mole of an ideal gas in a reversible isothermal expansion shown in the figure. He by mistake calculated the area of the shaded area in the PV graph as work and answered the magnitude of work equal to 49.26 L-atm. What is the correct magnitude of work done by the gas, in l-atm? (R = 0.0821 L-atm/K-mol)

TestHub · Accepted Answer

The shaded rectangle runs from V = 0 to the final volume V 2 = 4 L at the final pressure, so its area is P 2 V 2 = nRT = 49.26 L-atm (this gives T = 49.26/(2 × 0.0821) = 300 K). The correct work for the isothermal expansion from 2 L to 4 L is W = nRTln(V 2 )/(V 1 ) = 49.26 × ln 2 = 49.26 × 0.693 = 34.14 L-atm. Matches (d).

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