A process $1 \to 2$ using monoatomic gas is shown on the P - V diagram on the right. $P_{1} = 2 P_{2} = 10^{6} {N/ m}^{2}, V_{2} = 4 V_{1} = 0.4 m^{3}$ . The heat absorbed by the gas in this process in kilojoule will be :

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Solution

From work done
$W = \frac{1}{2} \times 3 V_{0} \times P_{0} + 3 V_{0} \times P_{0}$
$W = \frac{3}{2} P_{0} V_{0} + 3 V_{0} P_{0}$
$W = \frac{9}{2} P_{0} V_{0}$
$Δ U = n C_{v} Δ T = n (\frac{3 R}{2}) (T_{f} - T_{i})$
Where, $Δ U$ = change in internal energy
$T_{f}$ = Final temperature
$T_{i}$ = initial temperature
$C_{V}$ = Heat capacity (constant volume)
$Δ U = \frac{3}{2} n R (T_{f} - T_{i}) = \frac{3}{2} [P_{f} V_{f} - P_{i} V_{i}]$
$Δ U = \frac{3}{2} [4 P_{0} V_{0} - 2 P_{0} V_{0}]$
$Δ U = 3 P_{0} V_{0}$
$Δ Q = Δ U + W = \frac{9}{2} P_{0} V_{0} + 3 P_{0} V_{0} = \frac{15}{2} P_{0} V_{0}$

$P_{0} = \frac{10^{6}}{2}, V_{0} = 0.1$
$Δ Q = \frac{15}{2} \times \frac{10^{6}}{2} \times 0.1 = 375000 J$
$Δ Q = 375 kJ$

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