In a given process on an ideal gas, `dW=0 and dQlt0.` Then for the gas

Options

A. the temperature will decrease
B. the volume will increase
C. the pressure will remain constant
D. the temperature will increase

Answer

A

Explanation

(a) From the first law of thermodynamics
`dQ=dU+dW`
Here `dW=0 (given)`
`:. dQ=dU`
Now since `dQlt0 (given)`
`:. dQ is negative`
`rArrdU=-ve rArr dU decreases.`
`rArr` Temperature decreases.