Consider the same expansion,but this time against a constant external pressure of $1 \ atm$.
(N/A) The work done during expansion against a constant external pressure is given by the formula $w = -p_{ex} \Delta V$.
Assuming the change in volume $\Delta V$ is $8 \ L$,the work done is $w = -1 \ atm \times 8 \ L = -8 \ L \ atm$.
Since $q = -w$ for an adiabatic process or in the context of energy balance where heat absorbed equals work done,we have $q = 8 \ L \ atm$.
Assuming the change in volume $\Delta V$ is $8 \ L$,the work done is $w = -1 \ atm \times 8 \ L = -8 \ L \ atm$.
Since $q = -w$ for an adiabatic process or in the context of energy balance where heat absorbed equals work done,we have $q = 8 \ L \ atm$.
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