Question 447: Thermodynamics & KTG

Question

$'n'$ moles of an ideal gas undergoes a process $A$ $ 	o $ $B$ as shown in the figure. The maximum temperature of the gas during the process will be :
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l7zdfrjf/3f95c8e8-6dbf-48eb-8ef7-6bcb445be0cc/1d6e16c0-32ef-11ed-8cf6-c1445513adbd/file-1l7zdfrjg.png?format=png)

Accepted Answer

The equation for the line is
$P = {{ - {P_0}} \over {{V_0}}}V + 3P$
[slope $=$ ${{ - {P_0}} \over {{V_0}}},\,\,c = 3{P_0}$ ]
$P{V_0} + {P_0}V = 3{P_0}{V_0}\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)$
But $\,\,\,\,\,\,\,\,\,\,\,\,pv = nRT$
$	herefore$ $p = {{nRT} \over v}\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} ight)$
From $(i)$ & $(ii)$ ${{nRT} \over v}{V_0} + {P_0}V = 3{P_0}{V_0}$
$	herefore$ $nRT{V_0} + {P_0}{V^2} = 3{P_0}{V_0}\,\,\,\,\,...\left( {iii} ight)$
For temperature to be maximum ${{dT} \over {dv}} = 0$
Differentiating e.q.$(iii)$ by $'v'$ we get
$nR{V_0}{{dT} \over {dv}} + {P_0}\left( {2v} ight) = 3{P_0}{V_0}$
$	herefore$ $nR{V_0}{{dT} \over {dv}} = 3{P_0}{V_0} - 2{P_0}V$
${{dT} \over {dv}} = {{3{P_0}{V_0} - 2{P_0}V} \over {nR{V_0}}} = 0$
$V = {{3{V_0}} \over 2}$
$	herefore$ $p = {{3{P_0}} \over 2}\,\,\,\,\,\,\,\,\,$ [From $(i)$ ]
$	herefore$ ${T_{\max }} = {{9{P_0}{V_0}} \over {4nR}}\,\,\,\,\,\,\,\,\,\,$ [ From $(ii)$ ]
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l7zdgqjq/1421d9c1-c911-42d2-a2c2-dd68b18b6a54/38792770-32ef-11ed-8cf6-c1445513adbd/file-1l7zdgqjr.png?format=png)

Answer

${{9{P_0}{V_0}} \over {2nR}}$

Answer

${{9{P_0}{V_0}} \over {nR}}$

Answer

${{9{P_0}{V_0}} \over {4nR}}$

Answer

${{3{P_0}{V_0}} \over {2nR}}$

Thermodynamics & KTG

'n'$ moles of an ideal gas undergoes a process $A$ $ \to $ $B$ as shown in the figure. The maximum temperature of the gas during the process will be :

Choose the correct answer: