Ten moles of a certain ideal gas at a temperature 300 K was cooled isochorically so that the gas pressure reduced 2.0 times. Then, as a result of the isobaric process, the gas expanded till its temper

Question

Ten moles of a certain ideal gas at a temperature 300 K was cooled isochorically so that the gas pressure reduced 2.0 times. Then, as a result of the isobaric process, the gas expanded till its temperature got back to the initial value. The total amount of heat absorbed (in Kcal) by the gas in this process is

TestHub · Accepted Answer

Isochoric cooling at constant V: P halved ⇒ T₂ = 300/2 = 150 K. Isobaric reheating at P₁/2 from 150 K back to 300 K (ΔT = 150 K). Over the whole process the gas returns to 300 K, so ΔU t otal = 0 and net heat = net work. Only the isobaric step involves work: W = P₂ΔV = nRΔT. Q n et = nRΔT = 10 × 2 cal K⁻¹ mol⁻¹ × 150 K = 3000 cal = 3 kcal. (Independent of the gas's C v , which is why the gas need not be specified.)

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