A gas is allowed to expand in a well-insulate | vedclass

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(B) For an irreversible expansion process,the work done is given by $W = -P_{ext} \Delta V$. Given $P_{ext} = 2.5 \ atm$,$V_i = 2.50 \ L$,and $V_f = 4

Vedclass · Accepted Answer

$-505$

Vedclass · Answer

(B) For an irreversible expansion process,the work done is given by $W = -P_{ext} \Delta V$. Given $P_{ext} = 2.5 \ atm$,$V_i = 2.50 \ L$,and $V_f = 4.50 \ L$. $W = -2.5 \ atm 	imes (4.50 \ L - 2.50 \ L) = -2.5 	imes 2.0 = -5.0 \ L \ atm$. Since $1 \ L \ atm = 101.3 \ J$,$W = -5.0 	imes 101.3 \ J = -505 \ J$. As the container is well-insulated,the process is adiabatic,so $q = 0$. According to the First Law of Thermodynamics,$\Delta U = q + W$. Therefore,$\Delta U = 0 + (-505 \ J) = -505 \ J$.

$A$ gas is allowed to expand in a well-insulated container against a constant external pressure of $2.5 \ atm$ from an initial volume of $2.50 \ L$ to a final volume of $4.50 \ L$. The change in internal energy $\Delta U$ of the gas in joules will be .............. $J$.

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