A piston of volume 0.1 m3 contains two moles of a monatomic ideal gas at 300K. If it undergoes an isothermal process and expands until the internal pressure matches the external pressure P2 = 1atm. How much work is done by the gas on the environment?

the work is done by the gas on the environment - is W = - 3534.94 J (since the initial pressure is lower than the atmospheric pressure, it needs external work to expand)

Explanation:

assuming ideal gas behaviour of the gas, the equation for ideal gas is

P*V=n*R*T

where

P = absolute pressure

V = volume

T = absolute temperature

n = number of moles of gas

R = ideal gas constant = 8.314 J/mol K

P=n*R*T/V

the work that is done by the gas is calculated through

W=∫pdV = ∫ (n*R*T/V) dV

for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:

W=∫pdV = ∫ (n*R*T/V) dV = n*R*T ∫ (1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

P₂=n*R*T/V₂

dividing both equations

V₂/V₁ = P₁/P₂

W = n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂)

replacing values

P₁=n*R*T/V₁ = 2 moles * 8.314 J/mol K * 300K / 0.1 m3 = 49884 Pa

since P₂ = 1 atm = 101325 Pa

W = n*R*T * ln (P₁/P₂) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = - 3534.94 J