Answer to Question #263456 in Molecular Physics | Thermodynamics for den

Question

A closed system containing a gas expands slowly in a piston/cylinder from 600kPa and

0.1 m3

to a final volume of 0.5 m3

. Determine the work done if the pressure distribution

is determined to be: a) Pv = C, b) pV1.4 = C, c) P = C,

d) p = -300V + 630 kPa where V is in m3

Answer 1

(a) W.D.(W), if PV = C

$W = P_1V_1ln \frac{V_2}{V_1} = 600 \frac{kN}{m^2} \times 0.1 \;m^3 ln \frac{0.5}{0.1} = 96.57 \; kJ$

(b) W.D.(W), if PV^{1.4}=C

We know that,

$P_2=P_1 [\frac{V_1}{V_2}]^{1.4} = 600 [\frac{0.1}{0.5}]^{1.4} = 63.037 \; kPa \\ W = \frac{P_1V_1 -P_2V_2}{n-1} = \frac{(600 \times 0.1) -(630.04 \times 0.5)}{1.4-1} \frac{kN}{m^2}m^3 = 71.204 \; kJ$

(c) W.D.(W), if P=C

$W=P(V_2-V_1) = 600 \frac{kN}{m^2}(0.5-0.1)m^3 = 240 \;kJ$

(d) W.D.(W) if P = -300V+630

$W = \int^{V_2}_{V_1} PdV = \int^{V_2}_{V_1} (-300V+630)dV = [-150V^2]^{V_2}_{V_1} + [630V]^{V_2}_{V_1} \\ = -150(V_2^2 -V^2_1) +630(V_2- V_1) \\ = -150 (0.5^2 +0.1^2) + 630(0.5- 0.1) = 216 \;kJ$