ThermodynamicsHard

Question

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

Options

A.49.26
B.98.52
C.78.63
D.34.14

Solution

Area = P2 × ΔV ⇒ P2 × 4 = 49.26 L-atom

Now, correct work, $w = - nRT.\ln\frac{V_{2}}{V_{1}} = - P_{2}V_{2}.\ln\frac{4}{2}$= –49.26 × 0.693 = – 34.137 L-atom

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