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  4. 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

NTA AbhyasNTA Abhyas 2022

Solution:

From the first law of thermodynamics
dQ = dU + dW
Hence dW = 0 (given)
$\therefore $ dQ = dU
Now since dQ < 0 (given)
$\therefore dQ\text{is negative}\Rightarrow dU=-ve$
$\Rightarrow $ As U decreases, hence temperature decreases