Question

A 2400 cm3 container holds 0.14 mol of helium gas at 340 ∘C

a. How much work must be done to compress the gas to 1200 cm^3 at constant pressure?

b. How much work must be done to compress the gas to 1200 cm^3 at constant temperature?

Answer 1

A) Work done by an isobaric process is given by:

W = P*(V2 - V1)

To find the constant pressure, use the ideal gas law:

P*V1 = n*R*T1

Solve for P:

P = n*R*T1

Thus:

W = n*R*T1*(V2 - V1)/V1

B) Work done by an isothermal process is given by:

W = P1*V1*ln(V2/V1)

If you don't know what ln, as in natural logarithm, means...learn it!

And as like before, we can replace P1:

W = (n*R*T1/V1)*V1*ln(V2/V1)

Cancel the V1's:

W = n*R*T1*ln(V2/V1)

Summary:

A) W = n*R*T1*(V2 - V1)/V1

B) W = n*R*T1*ln(V2/V1)

Data: (volumes in cubic meters, temperature in Kelvin)

n:=0.14 moles; V1:= 2.4e-3 m^2; V2:=1.2e-3 m^2; T1:=613.15 K; R:=8.314 J/mole-K;

Results:

A) W = -362.66 Joules

B) W = 494.68 Joules