A sample of ideal gas is taken through the cyclic process as shown in the figure. The temperature of the gas at state A is $T_{A} = 200 K$ . At states B and C the temperature of the gas is the same. The greatest temperature of the gas in Kelvin during the cyclic process is $T_{\max} .$ Find $T_{\max} / 800.$

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Detailed Solution

At B and C $(T_{B} = T_{C})$
$\frac{P_{B} V_{B}}{T_{B}} = \frac{P_{C} V_{C}}{T_{C}}$
$\frac{P_{B} V_{A}}{T_{B}} = \frac{P_{C} V_{A}}{T_{C}}$
$P_{B} = 3 P_{C}$
For line $A C P α V$ (straight line through origin)
So $\frac{P_{C}}{P_{A}} = \frac{V_{C}}{V_{A}} = \frac{3 V_{A}}{V_{A}}$
$\Rightarrow P_{C} = 3 P_{A}$
Thus $P_{C} = 3 P_{A}; P_{B} = 3 P_{C} = 9 P_{A}$ …..(i)
$V_{C} = 3 V_{A}; V_{B} = V_{A}$
$T_{A} = \frac{P_{A} V_{A}}{n R}$ …..(ii)
From A to B ; sochoric $P ↑ T ↑$
So $T_{B} > T_{A}$
For C to A; both (P, V) $↓ s o T ↓$
Thus from B to C (we could have maximum temperature)
P = aV + b
$\Rightarrow P = - (\frac{6 P_{A}}{2 V_{A}}) + 12 P_{A}$ $\Rightarrow P = - \frac{3 P_{A} V}{V_{A}} + 12 P_{A}$
PV = nRT
$(- \frac{3 P_{A} V}{V_{A}} + 12 P_{A}) V = n R T$
For $T_{\max} \frac{d T}{d V} = 0$
$\frac{- 6 P_{A}}{V_{A}} V + 12 P_{A} = 0$
$V = 2 V_{A} \Rightarrow P = 6 P_{A}$
$T = \frac{6 P_{A} (2 V_{A})}{n R} = \frac{12 P_{A} V_{A}}{n R}$

$T_{\max} = 12 T_{A} = 2400 K$