A gas occupies 1.5 L at S.T.P. It is provide | vedclass

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(A) According to the first law of thermodynamics,$\Delta U = q + w$. Here,$q = +300 \ J$ (heat is provided to the system). The work done $w = -P_{ext}

Vedclass · Accepted Answer

$249.5$

Vedclass · Answer

(A) According to the first law of thermodynamics,$\Delta U = q + w$. Here,$q = +300 \ J$ (heat is provided to the system). The work done $w = -P_{ext} \Delta V$. Given $P_{ext} = 1 \ atm = 101.3 \ kPa$ and $\Delta V = (2 - 1.5) \ L = 0.5 \ L$. $w = -1 \ atm 	imes 0.5 \ L = -0.5 \ L \cdot atm$. Since $1 \ L \cdot atm = 101.3 \ J$,$w = -0.5 	imes 101.3 \ J = -50.65 \ J$. Therefore,$\Delta U = 300 \ J - 50.65 \ J = 249.35 \ J$. Rounding to the nearest provided option,the value is $249.5 \ J$.

$A$ gas occupies $1.5 \ L$ at $S.T.P.$ It is provided with $300 \ J$ of heat so that its volume becomes $2 \ L$ at a pressure of $1 \ atm$. The value of $\Delta U$ for the process is.....joule.

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