1 mole of an ideal gas, initially at 400 K and 10 atm is first expanded at constant pressure till the volume is doubled. Then the gas is made to undergo an isochoric process, in which its temperature is found to decrease. In the last step, the gas was compressed reversibly and adiabatically to its initial state. Determine the net work involved in this cyclic process (in terms of R). Given, $C_{V}$ for gas $= 1.5 R, (4)^{- 1 / 3} = 0.63$ . If $| W | = 2 R \times z$ , what is the value of $z$ ?

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Solution

(1) (10 atm, 400 K, 1 mole) $\frac{v}{d o u b l e s} \to v c o n s t a n t b u t T \to (3) A d i a b a t i c - ------------- \to r e v e r s i b l e c o m p r e s s i o n (4) \to s a m e a s (1)$
Now, $V_{(1)} = \frac{n R T_{(1)}}{P_{(1)}} = \frac{1 \times R \times 400}{10} = 40 R$
$V_{(2)} = 2 \times V_{(1)} = 80 R = V_{(3)}$
P - V curve is :
now, (3) $\to$ (1) : Adiabatic reversible.
So, $P_{(1)} V_{(1)} = P_{(3)} V_{(3)^{γ}} (h e r e, γ = \frac{5}{3})$
$∴ 10 \times (40 R)^{\frac{8}{3}} = P_{(3)} (80 R)^{\frac{5}{3}}$
$\Rightarrow P_{(3)} = 10 \times {(\frac{40 R}{80 R})}^{\frac{5}{3}} = \frac{10}{2^{\frac{5}{3}}} = \frac{5}{4^{\frac{1}{3}}}$
or, $P_{(3)} = 5 \times 0.63 = 3.15 a t m .$
$ω_{t o t a l} = ω_{(1) \to (2)} + ω_{(2) \to (3)} + ω_{(3) \to (1)}$
$ω_{(1) \to (2)} = 10 a t m (80 R - 40 R) = - 400 R$
$ω_{(2) \to (3)} = z e r o (i s o c h o r i c)$
$ω_{(3) \to (1)} = △ E (a d i a b a t i c)$
$= \frac{P_{(1)} V_{(1)} - P_{(3)} V_{(3)}}{γ - 1} = \frac{3}{2} [400 R - 252 R]$
$= 222 R .$
$∴ ω_{t o t a l} = - 400 R + 222 R = - 178 R$
$∴ | ω | = 178 R = 2 R \times 89 = 2 R \times Z$
$∴ Z = 89$

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