Question

One mole of helium gas, initially at STP (p₁ = 1 atm, T₁ = 0°C), undergoes an isovolumetric process in which its pressure falls to half its initial value. The work done by the gas is:

• A. 101 J
• B. 51 J
• C. 23 J
• D. 0 J

Hint:

In an isovolumetric process, the volume remains constant, meaning no work is done by the gas.

Answer 1

The Correct Option is D

In this problem, we need to determine the work done by one mole of helium gas during an isovolumetric process.

First, let's understand the key term:
Isovolumetric Process: This is a thermodynamic process in which the volume remains constant. Since the volume does not change, no work is done by or on the gas during the process. Work done is given by the formula:
W = P∆V
where W is the work done, P is the pressure, and ∆V is the change in volume.

Since the process is isovolumetric (∆V = 0):
W = P * 0 = 0

Therefore, the work done by the gas is simply 0 J.