A given mass of a gas expands from a state A | vedclass

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(A) The work done by a gas in a process is given by the area under the $P-V$ curve. In a $T-V$ diagram,the work done is proportional to the area under

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$W_1 > W_2 > W_3$

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(A) The work done by a gas in a process is given by the area under the $P-V$ curve. In a $T-V$ diagram,the work done is proportional to the area under the curve if we consider the ideal gas equation $PV = nRT$,which implies $P = nRT/V$. Thus,$W = \int P \, dV = \int (nRT/V) \, dV$. For a given expansion from $V_A$ to $V_B$,the work done is greater if the temperature $T$ is higher throughout the process. Looking at the $T-V$ diagram: Path $1$ involves an increase in temperature at constant volume followed by an expansion at constant temperature (or higher average temperature). Path $2$ is a straight line from $A$ to $B$. Path $3$ involves an expansion at constant temperature (or lower average temperature) followed by an increase in temperature at constant volume. Comparing the areas under the curves in the $P-V$ plane (which corresponds to the integral of $T/V$ with respect to $V$),we find that the path that stays at higher temperatures for larger volumes results in more work. Thus,the work done follows the order $W_1 > W_2 > W_3$.

$A$ given mass of a gas expands from a state $A$ to the state $B$ by three paths $1, 2$ and $3$ as shown in the $T-V$ indicator diagram. If $W_1, W_2$ and $W_3$ respectively be the work done by the gas along the three paths,then

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