(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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Thermodynamics

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Mix Examples-Thermodynamics and Thermochemistry

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IIT 2006 Paper All IIT years →

For a reaction $2 CO_{(g)} + O_{2(g)} \rightleftharpoons 2 CO_{2(g)}$,$\Delta_{r} G^0 = -128 \ kJ$ at $300 \ K$. If $\Delta_{r} S^0$ of the reaction is $-40 \ J \ K^{-1}$,calculate $\Delta_{r} U$ of the reaction. (in $kJ$)

MediumAP EAMCET 2022

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Calculate the enthalpy change on freezing of $1.0 \ mol$ of water at $10.0^{\circ} C$ to ice at $-10.0^{\circ} C$. Given: $\Delta_{fus} H = 6.03 \ kJ \ mol^{-1}$ at $0^{\circ} C$,$C_p [H_2 O_{(l)}] = 75.3 \ J \ mol^{-1} \ K^{-1}$,$C_p [H_2 O_{(s)}] = 36.8 \ J \ mol^{-1} \ K^{-1}$.

Medium

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The heat liberated when $1.89 \ g$ of benzoic acid is burnt in a bomb calorimeter at $25^{\circ} C$ increases the temperature of $18.94 \ kg$ of water by $0.632^{\circ} C$. If the specific heat of water at $25^{\circ} C$ is $0.998 \ cal / (g^{\circ} C)$,then find the heat of combustion of benzoic acid.

MediumAP EAMCET 2021

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Match the conditions and the temperature for the 'fusion' process of $H_2O_{(s)}$ at $1 \ atm$ pressure.

Condition Temperature

$(1)$. Spontaneous $(p)$. $273 \ K$

$(2)$. At equilibrium $(q)$. $260 \ K$

$(3)$. Non-spontaneous $(r)$. $280 \ K$

Condition	Temperature
$(1)$. Spontaneous	$(p)$. $273 \ K$
$(2)$. At equilibrium	$(q)$. $260 \ K$
$(3)$. Non-spontaneous	$(r)$. $280 \ K$

Medium

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The $C_p$ of an ideal gas is $10.314 \ J \ mol^{-1} \ K^{-1}$. One mole of this gas is expanded against a constant pressure of $p \ atm$. The change in temperature during expansion is $1.0 \ K$. The values of $q$ (in $J$) and $\Delta H$ (in $J \ mol^{-1}$) are respectively

MediumTS EAMCET 2025

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For a reaction $2 CO_{(g)} + O_{2(g)} \rightleftharpoons 2 CO_{2(g)}$,$\Delta_{r} G^0 = -128 \ kJ$ at $300 \ K$. If $\Delta_{r} S^0$ of the reaction is $-40 \ J \ K^{-1}$,calculate $\Delta_{r} U$ of the reaction. (in $kJ$)

Calculate the enthalpy change on freezing of $1.0 \ mol$ of water at $10.0^{\circ} C$ to ice at $-10.0^{\circ} C$. Given: $\Delta_{fus} H = 6.03 \ kJ \ mol^{-1}$ at $0^{\circ} C$,$C_p [H_2 O_{(l)}] = 75.3 \ J \ mol^{-1} \ K^{-1}$,$C_p [H_2 O_{(s)}] = 36.8 \ J \ mol^{-1} \ K^{-1}$.

Match the conditions and the temperature for the 'fusion' process of $H_2O_{(s)}$ at $1 \ atm$ pressure. Condition Temperature $(1)$. Spontaneous $(p)$. $273 \ K$ $(2)$. At equilibrium $(q)$. $260 \ K$ $(3)$. Non-spontaneous $(r)$. $280 \ K$

The $C_p$ of an ideal gas is $10.314 \ J \ mol^{-1} \ K^{-1}$. One mole of this gas is expanded against a constant pressure of $p \ atm$. The change in temperature during expansion is $1.0 \ K$. The values of $q$ (in $J$) and $\Delta H$ (in $J \ mol^{-1}$) are respectively

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Match the conditions and the temperature for the 'fusion' process of $H_2O_{(s)}$ at $1 \ atm$ pressure.

Condition Temperature

$(1)$. Spontaneous $(p)$. $273 \ K$

$(2)$. At equilibrium $(q)$. $260 \ K$

$(3)$. Non-spontaneous $(r)$. $280 \ K$