For the reaction, 2 CO+O_2 arrow 2 CO_2 ; Δ 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

Correct answer: -557 array ll & Δ H=Δ U+Δ(P V) or & Δ H=Δ U+V Δ P or & Δ U=Δ H-V Δ P array Given that, Δ H=-560 kJ Δ P=40-70=-30 atm array rlrl & & V Δ P & =1 L ×-30 atm=-30 L atom & & 1 L -atom & =0.1 kJ & V Δ P & =-30 × 0.1 kJ & Δ U & =-560-(-30 × 0.1)=-557 kJ array

JEE Advanced 2006

Exam: JEE Advanced
Year: 2006
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Advanced 2006 · Chemistry · Thermodynamics (C)

For the reaction, 2 CO+O_2 arrow 2 CO_2 ; Δ H=-560 ~kJ . Two moles...

For the reaction,

2 \mathrm{CO}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2 ; \Delta H=-560 \mathrm{~kJ}

. Two moles of

\mathrm{CO}

and one mole of

\mathrm{O}_2

are taken in a container of volume

1 \mathrm{~L}

. They completely form two moles of

\mathrm{CO}_2

, the gases deviate appreciably from ideal behaviour. If the pressure in the vessel changes from 70 to

40 \mathrm{~atm}

, find the magnitude (absolute value)

of \Delta U

500 \mathrm{~K} .(1 \mathrm{~L}-\mathrm{atm}=0.1 \mathrm{~kJ})

Correct answer

-557

Step-by-step solution

\because

\begin{array}{ll} \because & \Delta H=\Delta U+\Delta(P V) \\ \text{ or } & \Delta H=\Delta U+V \Delta P \\ \text{ or } & \Delta U=\Delta H-V \Delta P \end{array}

Given that,

\Delta H=-560 \mathrm{ kJ} \Delta P=40-70=-30 \mathrm{ atm}

\begin{array}{rlrl} & \therefore & V \Delta P & =1 \mathrm{ L} \times-30 \mathrm{ atm}=-30 \mathrm{ L} \text{ atom } \\ & & 1 \mathrm{ L} \text{-atom } & =0.1 \mathrm{ kJ} \\ \therefore & V \Delta P & =-30 \times 0.1 \mathrm{ kJ} \\ \therefore & \Delta U & =-560-(-30 \times 0.1)=-557 \mathrm{ kJ} \end{array}

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