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(A) The work done during the expansion of a gas against a constant external pressure is given by the formula: $W = -P_{ext} 	imes \Delta V$. Here,$P_

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$-456$

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(A) The work done during the expansion of a gas against a constant external pressure is given by the formula: $W = -P_{ext} 	imes \Delta V$. Here,$P_{ext} = 1 \ atm$,$V_1 = 15.5 \ dm^3$,and $V_2 = 20 \ dm^3$. Change in volume,$\Delta V = V_2 - V_1 = 20 \ dm^3 - 15.5 \ dm^3 = 4.5 \ dm^3$. Since $1 \ dm^3 \cdot atm = 101.325 \ J$,the work done is $W = -1 \ atm 	imes 4.5 \ dm^3 = -4.5 \ dm^3 \cdot atm$. Converting to joules: $W = -4.5 	imes 101.325 \ J \approx -456 \ J$.

Calculate the work done in joule if $2 \ mol$ of an ideal gas expands isothermally from $15.5 \ dm^3$ to $20 \ dm^3$ at a constant external pressure of $1 \ atm$. (in $J$)

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