In an isobaric process (`lambda` is adiabatic exponent of the gas)

Options

A. The heat given to gas is `(lambda)/(lambda-1)` times the work done by gas.
B. The work done by gas is `(lambda-1)` time the change in internal energy.
C. The temperature of gas in increased.
D. The temperature of gas is decreased.

Explanation

At constant pressure.
If volume increases. Temperature also increases
volume decreases,
In isobaric process,
`DeltaQ=lambda.DeltaU`
`therefore DeltaQ=DeltaU+DeltaW`
`therefore DeltaW=lambda.DeltaU-DeltaU=(y-1)DeltaU`
`DeltaW=(y-1)DeltaU`
`thereforeDeltaQ=(lambda)/((y-1))DeltaW`