JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
\(2.2\, g\) of nitrous oxide \(\left( N _{2} O \right)\) gas is cooled at a constant pressure of \(1\, atm\) from \(310\, K\) to \(270 \,K\) causing the compression of the gas from \(217.1 \,mL\) to \(167.75 \,mL\). The change in internal energy of the process, \(\Delta U\) is ' \(- x ^{\prime} J\). The value of ' \(x\) ' is \(....\) [nearest integer] (Given: atomic mass of \(N =14\, g \,mol ^{-1}\) and of \(O\) \(=16\, g\, mol ^{-1}\). Molar heat capacity of \(N _{2} O\) is \(100\, JK ^{-1}\, mol ^{-1}\) )
- A \(455\)
- B \(45\)
- C \(95\)
- D \(195\)
Answer & Solution
Correct Answer
(D) \(195\)
Step-by-step Solution
Detailed explanation
\(N _{2} O\) moles \(=\frac{2.2}{44}=\frac{1}{20}\) \(\Delta H = nC _{ p } \Delta T =\frac{1}{20} \times 100(-40)=-200\, J\) \(\Delta U = q _{ p }+ w\) \(w =- P _{\text {ext. }} \Delta\, V\) \(W =-1 \frac{(167.75-217.1)}{1000} \times 101.3 \,J\) \(W =+5 \,J\)…
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