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Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in the figure.

a) find the work done when the gas is taken from state 1 to state 2

b) what is the ratio of temperature $\frac{T_{1}}{T_{2}}$ if $V_{2}=2V_{1}$

c) given the internal energy for one mole of gas at temperature T is $\left (\frac{3}{2} \right )RT$ , find the heat supplied to the gas when it is taken from state 1 to 2 with $V_2 = 2V_1$ .

Answers (1)

Let $P_{1}V_{1}^{\frac{1}{2}}=P_BV_B^{\frac{1}{2}}=K$

And $P=\frac{K}{V^{\frac{1}{2}}}$

Work done for process 1 to 2; $WD=\int_{V_{1}}^{V_{2}}PdV=\int_{V_{1}}^{V_{2}}\frac{K}{V^{\frac{1}{2}}}dV=K\left [ \frac{V^{\frac{1}{2}}}{\frac{1}{2}} \right ]=2K\left [ \sqrt{V_{2}}-\sqrt{V_{1}} \right ]$

WD from V₁to V₂=

$dW=2P_1V_1^{\frac{1}{2}}V_2-V_1=2P_2V_2^{\frac{1}{2}}V_2-V_1$

Equation of ideal gas PV=nRT

$T=\frac{PV}{nR}=\frac{K\sqrt{V}}{nR }$

$T_{1}=\frac{K\sqrt{V_{1}}}{nR }$ and $T_{2}=\frac{K\sqrt{V_{2}}}{nR }$

$\frac{T_1}{T_2}=\frac{\sqrt{V_1}}{\sqrt{V_2}}=\frac{1}{\sqrt{2}}$

(c )

$\Delta U=\frac{3}{2}R\Delta T=\frac{3}{2}R(T_{2}-T_{1})$

$\Delta U=\frac{3}{2}RT_1(\sqrt{2}-1)$

$dW=2P_1V_1\frac{1}{2}\left (\sqrt{V_2}-\sqrt{V_1} \right )$

$dW=2P_1V_1^{\frac{1}{2}}(\sqrt{2V_1}-\sqrt{V_1})$

$dW=2P_1V_1(\sqrt{2}-1) =2nRT_{1}(\sqrt{2}-1)$

$n=1$

$dW=2RT_1(\sqrt{2}-1)$

$dQ=dW+dU=2RT_{1}(\sqrt{2}-1)+\frac{3}{2}RT_{1}\left (\sqrt{2}-1 \right )$

$=(\sqrt{2}-1)RT_{1}\left (2+\frac{3}{2} \right )$

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infoexpert24

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