Answer to Question #136159 in Molecular Physics | Thermodynamics for Annie Stewart

Question

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.)

Answer 1

solution

P₁= $6\times10^5N/m^2$ V₁ $=0.2m^3$

P₂=$2\times10^5N/m^2$ v₂ $=0.6m^3$

(a)if p varies linearly as function of V then

$P=a+bV$

putting the value of P₁ ,V₁ and P₂ V₂ in above equation

$6\times10^5=a+b\cdot 0.2\\2\times10^5=a+b\cdot 0.6$

from above equation a=$8\times10^5$

b=$-10\times10^5$

so work done by gas can be calculated as

$W=\int pdv$

$W=\int_{0.2}^{0.6}(8\times10^5-10^6V)dV$

$W=1.6\times10^5J$

(b)if PV=c then work done can be given as

$W=P_2V_2ln\frac{V_2}{V_1}$

$W=2\times10^5\times0.6ln\frac{0.6}{0.2}$

$W=0.56\times10^5J$

(C)if pvⁿ=constant then

$P_1V_!^n=P_2V_2^n$

by putting the value of Paressure and volume

$6\times10^5\times0.3^n=2\times10^5\times0.6^n\\n=1.58$

when n=1.58 then workdone is given as

$W=\frac{P_2V_2-P_1V_1}{1-n}$

$W=\frac{(2\times10^5\times0.6)-(6\times10^5\times0.3)}{1-1.58}$

$W=1.03\times10^5J$

so work done during volume expantion from 0.3m^3 to 0.6m^3 is $1.03\times10^5J .$