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Work Done in Isothermal Reversible Process

Two moles of ...

Question

Two moles of helium gas ( $γ = 5 / 3$ ) are initially at a temperature of $27^{o} C$ and occupy a volume of $20 l i t r e$ . The gas is first expanded at constant pressure until the volume is doubled. It then undergoes adiabatic change until the temperature returns to its initial value. If total work done by the gas is $- x k c a l$ , then what is $x$ ?

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Solution

The PV diagram for the given changes is as:

For isobaric process AB,
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ (from Charle's law)
$\frac{20}{300} = \frac{40}{T_{2}},$
$∴ T_{2} = 600 K$
$w_{t} = w_{A B} + w_{B C}$
Now, $w_{A B} = - n R d T$ (Isobaric process)
$= - 2 \times 2 \times (600 - 300) = - 1200 c a l .$
$w_{B C} =$ for adiabatic process
$w = - n C_{v} △ T$
$= - 2 \times \frac{3}{2} \times R \times (600 - 300)$
$= - 1800 C a l$
$∴ w_{T} = - 1200 - 1800 = - 3000 C a l = - 3 k c a l .$

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