An ideal gas expands by 1.5 L against a cons | vedclass

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Question

(B) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$. Given: $P_{ext} = 2 \ atm$,$

Vedclass · Accepted Answer

$-303.9$

Vedclass · Answer

(B) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$. Given: $P_{ext} = 2 \ atm$,$\Delta V = 1.5 \ L$. Substituting the values: $W = -2 \ atm 	imes 1.5 \ L = -3 \ atm \cdot L$. Since $1 \ atm \cdot L = 101.325 \ J$,we have: $W = -3 	imes 101.325 \ J = -303.975 \ J$. Rounding to one decimal place,we get $W \cong -303.9 \ J$.

An ideal gas expands by $1.5 \ L$ against a constant external pressure of $2 \ atm$ at $298 \ K$. Calculate the work done (in $J$)?

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