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(D) For an adiabatic process in an insulated container,the heat exchange $q = 0$. The work done $W$ during expansion against a constant external press

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$-1013 \ J$

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(D) For an adiabatic process in an insulated container,the heat exchange $q = 0$. The work done $W$ during expansion against a constant external pressure is given by $W = -P_{ext} \Delta V$. Given $P_{ext} = 10 \ atm$,$V_i = 4.2 \ L$,and $V_f = 5.2 \ L$. $\Delta V = V_f - V_i = 5.2 \ L - 4.2 \ L = 1.0 \ L$. $W = -10 \ atm 	imes 1.0 \ L = -10 \ atm \cdot L$. According to the first law of thermodynamics,$\Delta E = q + W$. Since $q = 0$,$\Delta E = W = -10 \ atm \cdot L$. Using the conversion factor $1 \ atm \cdot L = 101.3 \ J$,we get $\Delta E = -10 	imes 101.3 \ J = -1013 \ J$.

$A$ gas is allowed to expand in an insulated container against a constant external pressure of $10 \ atm$ from an initial volume $4.2 \ L$ to $5.2 \ L$. The change in internal energy $\Delta E$ of the gas will be $:-$

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