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
(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 \\\\\n\nW = \\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} \\\\\n\n= -150(V_2^2 -V^2_1) +630(V_2- V_1) \\\\\n\n= -150 (0.5^2 +0.1^2) + 630(0.5- 0.1) = 216 \\;kJ"
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