One mole of a gas occupying 3 L volume is ex | vedclass

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(C) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \Delta V$. Given: $P_{ext} = 1 \ bar$,$V_1 = 3 \ L$

Vedclass · Accepted Answer

$-1.200 	imes 10^3 \ J$

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(C) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \Delta V$. Given: $P_{ext} = 1 \ bar$,$V_1 = 3 \ L$,$V_2 = 15 \ L$. Change in volume $\Delta V = V_2 - V_1 = 15 \ L - 3 \ L = 12 \ L$. Since $1 \ L \ bar = 100 \ J$,we have $\Delta V = 12 \ dm^3$. $W = -1 \ bar 	imes (15 \ L - 3 \ L) = -12 \ L \ bar$. Converting to Joules: $W = -12 \ L \ bar 	imes 100 \ \frac{J}{L \ bar} = -1200 \ J = -1.200 	imes 10^3 \ J$.

One mole of a gas occupying $3 \ L$ volume is expanded against a constant external pressure of $1 \ bar$ to a volume of $15 \ L$. Calculate the work done by the system.

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