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(A) The formula for work done during expansion against a constant external pressure is $W = -P_{ext} \Delta V$. Given: $W = -600 \ J$,$V_1 = 15 \ dm^{

Vedclass · Accepted Answer

$1.2$

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(A) The formula for work done during expansion against a constant external pressure is $W = -P_{ext} \Delta V$. Given: $W = -600 \ J$,$V_1 = 15 \ dm^{3}$,$V_2 = 20 \ dm^{3}$. Change in volume $\Delta V = V_2 - V_1 = 20 \ dm^{3} - 15 \ dm^{3} = 5 \ dm^{3}$. Since $1 \ dm^{3} = 1 \ L$ and $1 \ L \cdot bar = 100 \ J$,we convert the volume change to units compatible with Joules. $5 \ dm^{3} = 5 \ L$. Using $W = -P_{ext} \Delta V$,we have $-600 \ J = -P_{ext} 	imes (5 \ L)$. $P_{ext} = \frac{600 \ J}{5 \ L} = 120 \ J/L$. Since $100 \ J = 1 \ L \cdot bar$,then $120 \ J = 1.2 \ L \cdot bar$. Therefore,$P_{ext} = \frac{120 \ J}{5 \ L} = 24 \ J/L = 0.24 \ bar$. Wait,re-evaluating the conversion: $1 \ dm^{3} \cdot bar = 100 \ J$. $P_{ext} (bar) = \frac{-W (J)}{100 	imes \Delta V (dm^{3})} = \frac{600}{100 	imes 5} = \frac{600}{500} = 1.2 \ bar$. The correct answer is $1.2 \ bar$.

Calculate the constant external pressure required to expand $2$ moles of an ideal gas from volume $15 \ dm^{3}$ to $20 \ dm^{3}$ if the amount of work done is $-600 \ J$. (in $bar$)

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