In a given process on an ideal gas, dW = 0 and dQ < 0 . Then for the gas

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Q: In a given process on an ideal gas , dW = 0 and dQ < 0 . Then for the gas

(a) the temperature will decreases

(b) the volume will increases

(c) the pressure will remain constant

(c) the temperature will increases

Ans: (a)

Sol: According to First Law of Thermodynamics ,

dQ = dU + dW

dW = 0

⇒ dQ = dU = n CV dT

dQ < 0

i.e., n CV dT < 0 i.e., temperature falls

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