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(C) The work done during expansion against a variable pressure is given by $W = -\int_{V_1}^{V_2} P \, dV$. Substituting $P = \frac{20}{V}$,we get $W

Vedclass · Accepted Answer

$5065.8$

Vedclass · Answer

(C) The work done during expansion against a variable pressure is given by $W = -\int_{V_1}^{V_2} P \, dV$. Substituting $P = \frac{20}{V}$,we get $W = -\int_{1}^{10} \frac{20}{V} \, dV$. $W = -20 \ln(\frac{10}{1}) = -20 	imes 2.303 \log_{10}(10) = -46.06 \ L \cdot atm$. Converting work to Joules: $W = -46.06 	imes 101.3 \ J \approx -4665.88 \ J$. According to the first law of thermodynamics,$\Delta U = q + W$. Given $\Delta U = 400 \ J$,we have $400 = q - 4665.88$. Therefore,$q = 400 + 4665.88 = 5065.88 \ J$.

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