In a given process on an ideal gas, dW = 0 and dQ

Question

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

Tardigrade Admin · Accepted Answer

From the first law of thermodynamics dQ = dU + dW Hence dW = 0 (given) ∴ dQ = dU Now since dQ < 0 (given) ∴ dQ textis negative⇒ dU=-ve ⇒ As U decreases, hence temperature decreases

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

The temperature will decrease

The volume will increase

The pressure will remain constant

The temperature will increase

From the first law of thermodynamics
dQ = dU + dW
Hence dW = 0 (given)
$∴$ dQ = dU
Now since dQ < 0 (given)
$∴ d Q is negative \Rightarrow d U = - v e$
$\Rightarrow$ As U decreases, hence temperature decreases

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

The temperature will decrease

The volume will increase

The pressure will remain constant

The temperature will increase

From the first law of thermodynamics dQ = dU + dW Hence dW = 0 (given) ∴ dQ = dU Now since dQ < 0 (given) ∴dQis negative⇒dU=−ve ⇒ As U decreases, hence temperature decreases

From the first law of thermodynamics
dQ = dU + dW
Hence dW = 0 (given)
$∴$ dQ = dU
Now since dQ < 0 (given)
$∴ d Q is negative \Rightarrow d U = - v e$
$\Rightarrow$ As U decreases, hence temperature decreases