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(B) Given the internal energy $U = A P^2 V$.For an adiabatic process, the first law of thermodynamics states $dQ = dU + dW = 0$, which implies $dU = -

Vedclass · Accepted Answer

$(A P+1)^2 V = 	ext{constant}$

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(B) Given the internal energy $U = A P^2 V$.For an adiabatic process, the first law of thermodynamics states $dQ = dU + dW = 0$, which implies $dU = -dW = -P dV$.Thus, $dU = -P dV$.Taking the differential of $U$ with respect to $V$: $dU = A P^2 dV + 2 A P V dP$.Equating the two expressions for $dU$: $-P dV = A P^2 dV + 2 A P V dP$.Dividing by $P$ (assuming $P 
eq 0$): $-dV = A P dV + 2 A V dP$.Rearranging the terms: $-dV - A P dV = 2 A V dP \Rightarrow -(1 + A P) dV = 2 A V dP$.Separating the variables: $-\frac{dV}{V} = \frac{2 A dP}{1 + A P}$.Integrating both sides: $-\int \frac{dV}{V} = \int \frac{2 A dP}{1 + A P}$.$-\ln V = 2 \ln(1 + A P) + C$.This simplifies to $\ln V + \ln(1 + A P)^2 = 	ext{constant}$.Therefore, $V(1 + A P)^2 = 	ext{constant}$.

Consider a thermodynamic process where internal energy $U = A P^2 V$ $(A = \text{constant})$. If the process is performed adiabatically, then:

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