One mole of an ideal gas undergoes a process | vedclass

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(B) Given the process equation $P = \frac{2V^2}{1+V^2}$.For $V_i = 1 \, m^3$,the initial pressure $P_i = \frac{2(1)^2}{1+(1)^2} = \frac{2}{2} = 1 \, P

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$+\frac{11}{5R} \, K$

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(B) Given the process equation $P = \frac{2V^2}{1+V^2}$.For $V_i = 1 \, m^3$,the initial pressure $P_i = \frac{2(1)^2}{1+(1)^2} = \frac{2}{2} = 1 \, Pa$.For $V_f = 2 \, m^3$,the final pressure $P_f = \frac{2(2)^2}{1+(2)^2} = \frac{8}{5} \, Pa$.Using the ideal gas equation $PV = nRT$,for $n = 1$ mole,$T = \frac{PV}{R}$.The change in temperature is $\Delta T = T_f - T_i = \frac{P_f V_f - P_i V_i}{nR}$.Substituting the values: $\Delta T = \frac{(\frac{8}{5} 	imes 2) - (1 	imes 1)}{1 	imes R} = \frac{\frac{16}{5} - 1}{R} = \frac{16-5}{5R} = \frac{11}{5R} \, K$.

One mole of an ideal gas undergoes a process $P = \frac{2V^2}{1+V^2} \, Pa$. The change in temperature of the gas when the volume changes from $V = 1 \, m^3$ to $V = 2 \, m^3$ is:

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