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(A) The enthalpy change $\Delta H$ is related to the internal energy change $\Delta U$ by the equation: $\Delta H = \Delta U + \Delta(PV)$.Given $\Del

Vedclass · Accepted Answer

$30$

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(A) The enthalpy change $\Delta H$ is related to the internal energy change $\Delta U$ by the equation: $\Delta H = \Delta U + \Delta(PV)$.Given $\Delta U = 35 \ 	ext{bar-L}$.The change in the product of pressure and volume is $\Delta(PV) = P_2V_2 - P_1V_1$.Substituting the given values: $P_1 = 2 \ 	ext{bar}, V_1 = 40 \ 	ext{L}, P_2 = 5 \ 	ext{bar}, V_2 = 15 \ 	ext{L}$.$\Delta(PV) = (5 \ 	ext{bar} 	imes 15 \ 	ext{L}) - (2 \ 	ext{bar} 	imes 40 \ 	ext{L}) = 75 \ 	ext{bar-L} - 80 \ 	ext{bar-L} = -5 \ 	ext{bar-L}$.Therefore, $\Delta H = 35 \ 	ext{bar-L} + (-5 \ 	ext{bar-L}) = 30 \ 	ext{bar-L}$.

List-$I$ (Process)	List-$II$ (Conditions)
$A$. Isothermal process	$I$. No heat exchange
$B$. Isochoric process	$II$. Carried out at constant temperature
$C$. Isobaric process	$III$. Carried out at constant volume
$D$. Adiabatic process	$IV$. Carried out at constant pressure

$2 \ \text{moles}$ of a gas undergoes a change in state from $(2 \ \text{bar}, 40 \ \text{L})$ to $(5 \ \text{bar}, 15 \ \text{L})$ at a constant temperature. If the change in internal energy is given by $35 \ \text{bar-L}$, then the enthalpy change for the process will be: (in $\text{bar-L}$)

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