Two moles of an ideal gas at temperature $T_{o} = 300 K$ was cooled isochorically so that the pressure was reduced to half. Then, in an isobaric process, the gas expanded till its temperature got back to the initial value. The total amount of heat absorbed by the gas in the process is:

2490 J

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2680 J

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240 J

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290 J

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Solution

The correct option is A 2490 J
Let the initial thermodynamic coordinates be ( $P_{0}, V_{0}, T_{0}$ ),followed by ( $P_{1}, V_{1}, T_{1}$ ) and finally ( $P_{2}, V_{2}, T_{2}$ )
Given $T_{0} = 300 K, P_{1} = \frac{P_{0}}{2}$
First process being isochoric, $V_{1} = V_{0}$
Thus, $T_{1} = \frac{T_{0}}{2} = 150 K$
Also, $T_{2} = T_{0} = 2 T_{1} = 300 K$
So, heat released in isochoric cooling= $Δ Q_{1} = n C_{v} Δ T$
and heat absorbed in isobaric heating= $Δ Q_{2} = n C_{p} Δ T$
Net heat absorbed= $Δ Q_{2} - Δ Q_{1} = n (C_{p} - C_{v}) Δ T = n R Δ T = 2 \times 8.3 \times 150 J = 2490 J$

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