In a given process on an ideal gas, $$dW = 0$$ and $$dQ < 0.$$ Then for the gas
A.
the temperature will decrease
B.
the volume will increase
C.
the pressure will remain constant
D.
the temperature will increase
Answer :
the temperature will decrease
Solution :
From the first law of thermodynamics
$$dQ = dU + dW$$
Here $$dW = 0$$ (given)
∴ $$dQ = dU$$
Now since $$dQ < 0$$ (given)
∴ $$dQ$$ is negative
⇒ $$dU = - ve$$
⇒ $$dU$$ decreases.
⇒ Temperature decreases.
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