JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
One mole of an ideal gas expands isothermally and reversibly from \(10 \mathrm{dm}^3\) to \(20 \mathrm{dm}^3\) at \(300 \mathrm{~K}. \Delta \mathrm{U}\), q and work done in the process respectively are :
Given : \(\mathrm{R}=8.3 \mathrm{JK}^{-1}\) and \(\mathrm{mol}^{-1}\)
\(\begin{aligned}
& \text { In } 10=2.3 \\
& \log 2=0.30 \\
& \log 3=0.48
\end{aligned}\)
- A \(0,21.84 \mathrm{~kJ},-1.26 \mathrm{~kJ}\)
- B \(0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}\)
- C \(0,21.84 \mathrm{~kJ}, 21,84 \mathrm{~kJ}\)
- D \(0,1.78 \mathrm{~kJ},-1.718 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(D) \(0,1.78 \mathrm{~kJ},-1.718 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & (10 \mathrm{~L}, 300 \mathrm{~K}) \xrightarrow{\mathrm{n}=1}(20 \mathrm{~L}, 300 \mathrm{~K}) \\ & -\mathrm{q}=\mathrm{w}=-\mathrm{nRT} \ln \frac{\mathrm{V}_2}{\mathrm{~V}_1} \\ & =-8.3 \times 300 \times \ln \left(\frac{20}{10}\right) \\ & =-1.718 \mathrm{~kJ}…
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