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(D) The process shown in the $PV$ diagram is an isothermal process because the curve follows the relation $PV = 	ext{constant}$.The work done in an i

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$17258$

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(D) The process shown in the $PV$ diagram is an isothermal process because the curve follows the relation $PV = 	ext{constant}$.The work done in an isothermal process is given by the formula:$W = nRT \ln \left( \frac{V_2}{V_1} ight)$Given values:$n = 1 \, 	ext{mole}$$T = 27^{\circ} {C} = 27 + 273 = 300 \, {K}$$R = 8.3 \, {J} / 	ext{mole} \cdot {K}$$V_1 = 2 \, {m}^3, V_2 = 4 \, {m}^3$$\ln 2 = 0.6931$Substituting the values:$W = 1 	imes 8.3 	imes 300 	imes \ln \left( \frac{4}{2} ight)$$W = 2490 	imes \ln 2$$W = 2490 	imes 0.6931$$W = 1725.819 \, {J}$To express this in the form $...... 	imes 10^{-1} \, {J}$:$W = 17258.19 	imes 10^{-1} \, {J}$Rounding off to the nearest integer,we get $17258 	imes 10^{-1} \, {J}$.

One mole of an ideal gas at $27^{\circ} {C}$ is taken from $A$ to $B$ as shown in the given $PV$ indicator diagram. The work done by the system will be $...... \times 10^{-1} \, {J}$.
[Given: $R = 8.3 \, {J} / \text{mole} \cdot {K}, \ln 2 = 0.6931$] (Round off to the nearest integer)

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One mole of an ideal gas at $27^{\circ} {C}$ is taken from $A$ to $B$ as shown in the given $PV$ indicator diagram. The work done by the system will be $...... \times 10^{-1} \, {J}$.[Given: $R = 8.3 \, {J} / \text{mole} \cdot {K}, \ln 2 = 0.6931$] (Round off to the nearest integer)