A gas occupies a volume of 2 L at S.T.P. It | vedclass

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(B) According to the first law of thermodynamics,$\Delta U = q + w$. Given: $q = +300 \ J$ (heat absorbed). Work done $w = -P_{ext} \Delta V$. $P_{ext

Vedclass · Accepted Answer

$249.5$

Vedclass · Answer

(B) According to the first law of thermodynamics,$\Delta U = q + w$. Given: $q = +300 \ J$ (heat absorbed). Work done $w = -P_{ext} \Delta V$. $P_{ext} = 1 \ atm = 1.01325 	imes 10^5 \ Pa$. $\Delta V = V_f - V_i = 2.5 \ L - 2.0 \ L = 0.5 \ L = 0.5 	imes 10^{-3} \ m^3$. $w = -(1.01325 	imes 10^5 \ Pa) 	imes (0.5 	imes 10^{-3} \ m^3) = -50.66 \ J$. $\Delta U = 300 \ J - 50.66 \ J = 249.34 \ J$. Rounding to the nearest provided option,the value is approximately $249.5 \ J$.

$A$ gas occupies a volume of $2 \ L$ at $S.T.P.$ It absorbs $300 \ J$ of heat. If the pressure is $1 \ atm$,the volume becomes $2.5 \ L$. The change in internal energy $(\Delta U)$ for the process in $Joule$ is:

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