Question

3. A sample of argon of mass 3.56 g occupies 12.5 dm at 305 K. (a) Calculate the work done when the gas expands isothermally against a constant external pressure of 70 kPa utilit volume has increased by 3.5 dm. (b) Calculate the work that would be done if the same expansion occurred reversibly.

Answer 1

(a)
Work done when the gas expands isothermally against a constant pressure,

$W=-p_{ext} \cdot \Delta V$

where, $p_{ext}$ = the constant external pressure = 7.0 kPa and $\Delta V$ = change in volume during the process = 3.5 dm³.

Hence, the work done here is

$W=-(7 \times 10^3 \ Pa) \times (3.5 \ dm^3)$

$W=-24500 \ Pa \ dm^3$

$W=-24.5 \ Pa \ m^3$

$W=-24.5 \ \frac{J}{m^3} \ m^3 \ \ \ \ \ \ \ \ \ \ \ (as \ 1\ Pa=1\ \frac{J}{m^3})$

$W=-24.5 \ J$

(b)

But if the gas expands in isothermal reversible way, the work done would be,

$W=-nRT\ln\frac{V_2}{V_1}$

where, $n$ = the number of moles = 3.56/39.95 = 0.089 moles

$R$ = universal gas constant = 8.314 J mol^-1 K^-1

$T$ = temperature = 305 K

V₁ = initial volume = 12.5 dm³

V₂ = final volume = (12.5 + 3.5) = 16 dm³

Hence, the work done would be,

$W=-0.089 \ moles \times (8.314 \ J \ mol^{-1} \ K^{-1})\times (305 \ K) \ln\frac{16 \ dm^3}{12.5 \ dm^3}$

$W=-55.71 \ J$