Method 5 of W
Heat taken from a gas in a process is 80 J and increase in internal energy of the gas is 20J. Find work done by the gas in the given process.

Answer

A

Explanation

Heat is taken from the gas.
Therefore, Q is negative. Or, `Q=-80J` Internal energy of the gas is increasing.
Therefore, `DeltaU` is positive. Or `DeltaU=+20J`
Using the first law eqution,
`Q=W+DeltaU` or `W=Q-DeltaU=-80-20=-100J`
Here, negative sign indicates that volume of the gas is decreasing and work is done on the gas.