(A) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta(PV)$.Since the volume $V$ is constan

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(A) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta(PV)$.Since the volume $V$ is constant,$\Delta(PV) = V \Delta P$.Therefore,$\Delta U = \Delta H - V \Delta P$.Given $\Delta H = -560 \ kJ$,$V = 1 \ L$,and $\Delta P = P_{final} - P_{initial} = 40 - 70 = -30 \ atm$.Substituting the values: $\Delta U = -560 - (1 \ L \times (-30 \ atm))$.Using the conversion factor $1 \ L \ atm = 0.1 \ kJ$,we get $\Delta U = -560 - (-30 \times 0.1) = -560 + 3 = -557 \ kJ$.The magnitude (absolute value) of $\Delta U$ is $|-557| = 557 \ kJ$.

Class 11 Chemistry Thermodynamics Mix Examples-Thermodynamics and Thermochemistry

For the reaction,$2 CO + O_2 \longrightarrow 2 CO_2$; $\Delta H = -560 \ kJ$. Two moles of $CO$ and one mole of $O_2$ are taken in a container of volume $1 \ L$. They completely form two moles of $CO_2$. The gases deviate appreciably from ideal behavior. If the pressure in the vessel changes from $70 \ atm$ to $40 \ atm$,find the magnitude (absolute value) of $\Delta U$ at $500 \ K$. $(1 \ L \ atm = 0.1 \ kJ)$

IIT 2006 All IIT

A
$557$
B
$478$
C
$654$
D
$324$

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DifficultJEE MAIN 2026

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In an irreversible process taking place at constant $T$ and $P$ and in which only pressure-volume work is being done,the change in Gibbs free energy $(dG)$ and change in entropy $(dS)$ satisfy the criteria:

MediumAIEEE 2003

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For the combustion reaction $2C_8H_{18}(g) + 25O_2(g) \rightarrow 16CO_2(g) + 18H_2O(g)$,what are the signs of $\Delta H$,$\Delta S$,and $\Delta G$?

Calculate the work done for the following reaction at $27^{\circ}C$: $C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)$ $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$ (in $\text{ J}$)

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