5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If ${C_V} = 28{\rm{ }}\,\,J{K^{ - 1}}{\rm{ }}mo{l^{ - 1}},$ calculate $\Delta U{\rm{ }}$ and $\Delta pV$ for this process, $(R = 8.0{\rm{ }}\,J{K^{ - 1}}{\rm{ }}mo{l^{ - 1}})$

$\Delta U = 14{\rm{ }}\,kJ;\Delta (pV) = 4{\rm{ }}\,kJ$

$\Delta U = 2.8{\rm{ }}\,kJ;\Delta (pV) = 0.8{\rm{ }}\,kJ$

$\Delta U = 14{\rm{ }}\,J;\Delta (pV) = 0.8{\rm{ }}\,J$

$\Delta U = 14{\rm{ }}\,kJ;\Delta (pV) = 18{\rm{ }}\,kJ$

JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

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Application +4 | -1

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If

{C_V} = 28\mathrm{ }\,\,J{K^{ - 1}}\mathrm{ }mo{l^{ - 1}},

calculate

\Delta U\mathrm{ }

and

\Delta pV

for this process,

(R = 8.0\mathrm{ }\,J{K^{ - 1}}\mathrm{ }mo{l^{ - 1}})

View answer and explanation

Correct option: C

\Delta U = 14\mathrm{ }\,kJ;\Delta (pV) = 4\mathrm{ }\,kJ

The correct answer is

\Delta U = 14\mathrm{ }\,kJ;\Delta (pV) = 4\mathrm{ }\,kJ

. The change in internal energy and the change in pV are calculated using the given values of

C_V

, R, and the temperature change.

Practice All Chapter MCQs →

\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{H}_3=\mathrm{c}

0.18 \mathrm{~g}

4 \mathrm{~K}.

\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right)

{CsI}

{CsCl}

{CsCl}

{Cs} I

\begin{array}{cccccccccccccccccccc} {}&{\Delta H(KJ/mol)} \\ {1/2A \to B}&{ + 150} \\ {3B \to 2C + D}&{ - 125} \\ {E + A \to 2D}&{ + 350} \end{array}

{B}+{D} \rightarrow {E}+2 {C}, \Delta {H}

\Delta H

\Delta u

\mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right) + 2{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 890.3\mathrm{kJ}