In an isovolumetric process where the pressure increases, work done on the gas, and the change in internal energy of the system positive, negative, or zero?

Question

W is +, and ΔU is +

W is –, and ΔU is 0

W is 0, and ΔU is +

W is 0, and ΔU is –

those are the choices given, please help me! thank you

Answer 1

Typically in such problems in thermodynamics we are dealing with pressure–volume work. This is defined under a change in volume:

dW = P * dV

We have a constant volume, so our pressure-volume work = 0

The wording of your question is not 100% clear, but it is clear you state the pressure increases. Increasing pressure in an isochoric- or isovolumetric-process means that heat must be added to the system to increase the pressure (see ideal-gas-law).

Since (delta)U, the change in energy is defined as:

(delta)U = Q - W

and since we now know W = 0, and Q is positive (heat added to system), U is positive.