Two moles of helium gas are taken over the cy | vedclass

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(A) The process $DA$ is an isochoric process because the temperature $T$ is constant at $300 	ext{ K}$ while the pressure changes from $1 	imes 10^5

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$+414 R$

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(A) The process $DA$ is an isochoric process because the temperature $T$ is constant at $300 	ext{ K}$ while the pressure changes from $1 	imes 10^5 	ext{ Pa}$ to $2 	imes 10^5 	ext{ Pa}$.However,looking at the $P-T$ diagram,the line $DA$ is vertical,meaning it is an isothermal process at $T = 300 	ext{ K}$.For an isothermal process,the work done by the gas is given by $W_{	ext{by}} = nRT \ln\left(\frac{V_f}{V_i}ight)$.Using the ideal gas law $PV = nRT$,we have $V = \frac{nRT}{P}$,so $\frac{V_f}{V_i} = \frac{P_i}{P_f}$.Thus,$W_{	ext{by}} = nRT \ln\left(\frac{P_D}{P_A}ight) = 2.303 nRT \log_{10}\left(\frac{P_D}{P_A}ight)$.Given $n = 2$,$T = 300 	ext{ K}$,$P_D = 1 	imes 10^5 	ext{ Pa}$,and $P_A = 2 	imes 10^5 	ext{ Pa}$.$W_{	ext{by}} = 2.303 	imes 2 	imes R 	imes 300 	imes \log_{10}\left(\frac{1 	imes 10^5}{2 	imes 10^5}ight) = 2.303 	imes 600 	imes R 	imes \log_{10}(0.5)$.$W_{	ext{by}} = 1381.8 	imes R 	imes (-0.301) \approx -416 R$ (using $\log_{10}(0.5) \approx -0.301$).Using the standard approximation $\log_{10}(0.5) \approx -0.3$,$W_{	ext{by}} = 2.303 	imes 600 	imes R 	imes (-0.3) = -414.54 R \approx -414 R$.The work done $ON$ the gas is $W_{	ext{on}} = -W_{	ext{by}} = -(-414 R) = +414 R$.

Two moles of helium gas are taken over the cycle $ABCDA$,as shown in the $P-T$ diagram. The work done on the gas in taking it from $D$ to $A$ is

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