The gas absorbs 100 J and is simultaneously | vedclass

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(C) Given: $q = +100 \ J$ (Heat absorbed). $P_{ext} = 1.50 \ atm$. $V_1 = 8 \ L, V_2 = 2 \ L$. $\Delta V = V_2 - V_1 = 2 - 8 = -6 \ L$. Work done $(w)

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$1011$

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(C) Given: $q = +100 \ J$ (Heat absorbed). $P_{ext} = 1.50 \ atm$. $V_1 = 8 \ L, V_2 = 2 \ L$. $\Delta V = V_2 - V_1 = 2 - 8 = -6 \ L$. Work done $(w)$ $= -P_{ext} \Delta V = -(1.50 \ atm) 	imes (-6 \ L) = 9.0 \ L \cdot atm$. Since $1 \ L \cdot atm = 101.325 \ J$,$w = 9.0 	imes 101.325 \ J = 911.925 \ J$. According to the first law of thermodynamics: $\Delta U = q + w$. $\Delta U = 100 \ J + 911.925 \ J = 1011.925 \ J \approx 1011 \ J$.

The gas absorbs $100 \ J$ and is simultaneously compressed by a constant external pressure of $1.50 \ atm$ from $8 \ L$ to $2 \ L$ volume. Hence $\Delta U$ will be ..... $J$.

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