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 :

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 $$(ii)$$ ]

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 :

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