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(D) For an adiabatic process,the heat exchange $Q = 0$. According to the first law of thermodynamics,$\Delta U = Q + W$. Since $Q = 0$,$\Delta U = W$.

Vedclass · Accepted Answer

$-835$

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(D) For an adiabatic process,the heat exchange $Q = 0$. According to the first law of thermodynamics,$\Delta U = Q + W$. Since $Q = 0$,$\Delta U = W$. The work done against a constant external pressure is given by $W = -P_{ext} \Delta V$. Given $P_{ext} = 0.75 \, atm$,$V_1 = 1 \, L$,and $V_2 = 12 \, L$. $\Delta V = V_2 - V_1 = 12 \, L - 1 \, L = 11 \, L$. $W = -0.75 \, atm 	imes 11 \, L = -8.25 \, atm \cdot L$. Converting $atm \cdot L$ to Joules $(1 \, atm \cdot L = 101.325 \, J)$: $\Delta U = -8.25 	imes 101.325 \, J = -835.93 \, J$. Thus,the change in internal energy is approximately $-835.9 \, J$.

If a gas expands adiabatically from $1 \, L$ to $12 \, L$ against a constant external pressure of $0.75 \, atm$,the $\Delta U$ of the gas is .... $J$. (in $.9$)

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