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JEE Advanced · Chemistry · 6. Thermodynamics (C)

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

  1. A -557
  2. B -667
  3. C -645
  4. D -145
Verified Solution

Answer & Solution

Correct Answer

(A) -557

Step-by-step Solution

Detailed explanation

\(\because \Delta H=\Delta U+\Delta(P V) \)
\( \text {or } \Delta H=\Delta U+V \Delta P \)
\( \text {or } \Delta U=\Delta H-V \Delta P\)
Given that, \(\Delta H=-560 \mathrm{~kJ}\)
\(
\Delta P=40-70=-30 \mathrm{~atm}
\)
\(\therefore V \Delta P =1 \mathrm{~L} \times-30 \mathrm{~atm}=-30 \mathrm{~L} \text { atom } \)
\( 1 \mathrm{~L} \text {-atom } =0.1 \mathrm{~kJ} \)
\( \therefore V \Delta P =-30 \times 0.1 \mathrm{~kJ} \)
\( \therefore \Delta U =-560-(-30 \times 0.1)=-557 \mathrm{~kJ}\)
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