Two moles of helium are initially at 23 $\circ$ C and 1 atm.The work done by the gas at constant pressure when the volume is doubled is

Question

Quizplus · Accepted Answer

The work done at constant pressure can be given by the formula:
W = -PΔV
where ΔV is the change in volume and P is the pressure.
Initially, the volume of 2 moles of helium is V1 = nRT/P = (2 mol)(0.0821 L atm/K mol)(23°C + 273.15)/1 atm = 98.96 L
When the volume is doubled, the new volume is V2 = 2V1 = 197.92 L
Therefore, the change in volume is ΔV = V2 - V1 = 98.96 L
Since the pressure is constant at 1 atm, the work done is:
W = -PΔV = -(1 atm)(98.96 L) = -98.96 L atm
Converting to joules using the conversion factor 1 L atm = 101.325 J, we get:
W = -98.96 L atm x 101.325 J/L atm = -10020.9 J
Rounding to two significant figures, we get the answer:
W = -4.92 kJ (since the work done is negative, we take the absolute value and express it in kilojoules)

Two Moles of Helium Are Initially at 23 ∘\circ∘ C and 1 Atm

Two Moles of Helium Are Initially at 23^$\circ$C and 1 Atm