Question

A 7.28 g sample of He had a volume of 25.2 dm3 at 315 K. A. Calculate the work done when the sample expands isothermally against a constant external pressure of 0.800 bar until the pressure is equalized. B. What would be the amount of work done if this expansion had occurred reversibly? (PLZ SHOW BOTH PARTS)

Answer 1

Weight /Mass of the sample: 7.28 g

n= Moles of gas= 7.28/4= 1.82

Volume of Gas: 25.2 dm³ = 25.2 L

Temperature: 315K

R= 0.0821 L atm K⁻¹ mol⁻¹ = 8.314 4621 × 10⁻² L bar K⁻¹ mol⁻¹

PV= nRT

P×25.2 = 1.82 x 8.314 4621 × 10⁻² x 315

P1 = 189.14× 10⁻² = 1.89 Bar

P2= 0.800 bar

Work done= - nRT [ln (P1/P2)]

Work done= - 1.82 x8.314 4621×315 × 10⁻² ln(1.89/0.800)

Work done= - 47.66× ln(1.89/0.800)

Work done= - J = - 40.97×10^-3 KJ

Answer B)

As work done in reversible isothermal expansion is given by:

w= -nRT ln (V₂/ V₁)     

Since the process is Isothermal, ∆U=0 and ∆H = 0 (as for an isothermal expansion of an ideal gas ∆U=0 and ∆H = 0)

V1= Volume of Gas: 25.2 dm³ = 25.2 L

Calculate V2 from work done in question one V2= 28.68

Work done= - 1.82 x8.314 4621×315 × ln(28.68/25.2) =- 616.56 J = -616.56 ×10^-3 KJ

Work done= From first law, ∆U= q + w
Since ∆U=0, q = - w = 616.56 ×10^-3 KJ

Work done in isothermal reversible expansion is 616.56 ×10^-3 KJ