In a process, the pressure of an ideal gas is proportional to square of the volume of the gas. If the temperature of the gas increases in this process, then work done by this gas

Options

A. is positive
B. is negative
C. is zero
D. may be positive

Answer

B

Explanation

`P prop 1/(V^2) implies P = k/(V^2) V implies PV^(2) = k`
`implies PV.V = k implies nRTV = k implies TV = k_(1)`
Since temperature increases therefore volume decreases.
Hence the work done by the gas should be negative.