Two kg of a substance undergoes the specified process in a cylinder–piston
arrangement :
p1 = 6 bar; V1 = 0.2 m3; p2 = 2 bar; V2 = 0.6 m3. Determine the work done
in each case (i) p varies as a linear function of V, (ii) pV = constant and
(iii) p remains constant till the volume reaches 0.3 m3; pVn = constant after
that.
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Answer to Question #136159 in Molecular Physics | Thermodynamics for Annie Stewart
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Molecular Physics | Thermodynamics
Question #136159
Two kg of a substance undergoes the specified process in a Cylinder-piston arrangement:
p1 = 6 bar; V1 = 0.2 m³; p2 = 2 bar; V2 = 0.6 m³,
Determine the work done (W) in each case.
(a) p varies as a linear function of V. (b) pV = a constant. (C) p remains constant till the volume reaches V3 = 0.3m³; pV^n = a Constant after that4 (You have to determine the value of n.)
Expert's answer
solution
P1= 6\times10^5N/m^26×105N/m2 V1 =0.2m^3=0.2m3
P2=2\times10^5N/m^22×105N/m2 v2 =0.6m^3=0.6m3
(a)if p varies linearly as function of V then
P=a+bVP=a+bV
putting the value of P1 ,V1 and P2 V2 in above equation
6\times10^5=a+b\cdot 0.2\\2\times10^5=a+b\cdot 0.66×105=a+b⋅0.22×105=a+b⋅0.6
from above equation a=8\times10^58×105
b=-10\times10^5−10×105
so work done by gas can be calculated as
W=\int pdvW=∫pdv
W=\int_{0.2}^{0.6}(8\times10^5-10^6V)dVW=∫0.20.6(8×105−106V)dV
W=1.6\times10^5JW=1.6×105J
(b)if PV=c then work done can be given as
W=P_2V_2ln\frac{V_2}{V_1}W=P2V2lnV1V2
W=2\times10^5\times0.6ln\frac{0.6}{0.2}W=2×105×0.6ln0.20.6
W=0.56\times10^5JW=0.56×105J
(C)if pvn=constant then
P_1V_!^n=P_2V_2^nP1V!n=P2V2n
by putting the value of Paressure and volume
6\times10^5\times0.3^n=2\times10^5\times0.6^n\\n=1.586×105×0.3n=2×105×0.6nn=1.58
when n=1.58 then workdone is given as
W=\frac{P_2V_2-P_1V_1}{1-n}W=1−nP2V2−P1V1
W=\frac{(2\times10^5\times0.6)-(6\times10^5\times0.3)}{1-1.58}W=1−1.58(2×105