Question

A 2000 cm^3 container holds 0.10 mol of helium gas at 300 °C. How much work must be done to compress the gas to 1000 cm^3 at constant pressure? How much work to compress the gas to 100 cm^3 at constant temperature?

Answer 1

Solution:

Given: Intial volume V_i = 2000cm³ = 2000*10^-6m³

final volume V_f = 1000cm³ = 1000*10^-6m³

Temperature T = 300 ^oC = (300+273)K = 573K

Number of moles n = 0.1 mol

Gas constant R = 8.31J/mol.K (standard data)

Part (I) First task is to find the intial pressure, for that we use the ideal gas law equation, PV=nRT

thus   P = nRT/V_i

P = (0.1mol)*( 8.31J/mol.K)*(573K)/(2000*10^-6m³)

P = 2.3808*10⁵ N/m²

To compress the gas at constant pressure indicates the isobaric pressure and the workdone W is given by

W = p(V_f -V_i)

W = (2.3808*10⁵ N/m²)*(1000*10^-6m³ - 2000*10^-6m³)

W = -238.08 J

Thus the workdone to compress the gas at constant pressure is 238.08J

Part (II)

final volume V_f = 100cm³ = 100*10^-6m³

Since the gas is compressed at constant temparature, it is isothermal process; the workdone by the gas is given by

W = nRT*ln(V_f/V_i)                   --------note that "ln" stands for natural log

W = (0.1mol)*(8.31J/mol.K)*(573K)*ln[100*10^-6m³ / 2000*10^-6m³)

W = -1426.5 J

Thus work done to compress the gas at constant temperature is 1426.5 J