Two moles of an ideal mono atomic gas undergoes compression from A to B as shown in the figure-T K -If for the process AB - P T 2- constant Then the work done in kcal on the system for the process AB is

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Compressibility Factor

Two moles of ...

Question

Two moles of an ideal mono atomic gas undergoes compression from A to B as shown in the figure:

If for the process AB, $P T^{2}$ = constant
Then the work done (in kcal) on the system for the process AB is

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Solution

From the graph,
$T_{1} = 1000 K$
$T_{2} = 500 K$
Given for the process $P T^{2} = C_{1}$ and we know PV = nRT
So $P {(\frac{P V}{n R})}^{2} = C_{1} \Rightarrow P^{3} V^{2} = C$
$\Rightarrow P = {(\frac{C}{V^{2}})}^{\frac{1}{3}}$
$W = - \int_{V_{1}}^{V_{2}} P d V = - \int_{V_{1}}^{V_{2}} C^{\frac{1}{3}} \frac{d V}{V^{\frac{2}{3}}}$
$= - C^{\frac{1}{3}} 3 (V_{2}^{\frac{1}{3}} - V_{1}^{\frac{1}{3}})$
$= - 3 (P_{2} V_{2} - P_{1} V_{1})$
$= 3 n R (T_{1} - T_{2})$
$= 3 \times 2 \times 2 \times 500$
$= 6000 C a l = 6 k C a l$

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Two moles of an ideal mono atomic gas undergoes compression from A to B as shown in the figure: If for the process AB, PT2 = constant Then the work done (in kcal) on the system for the process AB is

From the graph, T1=1000 K T2=500 K Given for the process PT2=C1 and we know PV = nRT So P(PVnR)2=C1⇒P3V2=C ⇒P=(CV2)13 W=−∫V2V1PdV=−∫V2V1C13dVV23 =−C133(V132−V131) =−3(P2V2−P1V1) =3nR(T1−T2) =3×2×2×500 =6000 Cal=6 kCal

Two moles of an ideal mono atomic gas undergoes compression from A to B as shown in the figure:

If for the process AB, $P T^{2}$ = constant
Then the work done (in kcal) on the system for the process AB is