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(D) For an insulated container,the process is adiabatic,so $q = 0$. According to the first law of thermodynamics,$\Delta U = q + w$. The work done dur

Vedclass · Accepted Answer

$-506.5$

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(D) For an insulated container,the process is adiabatic,so $q = 0$. According to the first law of thermodynamics,$\Delta U = q + w$. The work done during expansion against a constant external pressure is given by $w = -P_{ext} 	imes \Delta V$. Here,$P_{ext} = 2.5 \ atm$,$V_1 = 2.5 \ L$,and $V_2 = 4.5 \ L$. $\Delta V = V_2 - V_1 = 4.5 \ L - 2.5 \ L = 2.0 \ L$. $w = -2.5 \ atm 	imes 2.0 \ L = -5.0 \ atm \cdot L$. Since $1 \ atm \cdot L = 101.3 \ J$,we have $w = -5.0 	imes 101.3 \ J = -506.5 \ J$. Therefore,$\Delta U = 0 + (-506.5 \ J) = -506.5 \ J$.

$A$ gas is allowed to expand in an insulated container against a constant external pressure of $2.5 \ atm$ from $2.5 \ L$ to $4.5 \ L$. The change in internal energy of the gas in joules is: (in $J$)

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