KCET · Chemistry · Thermodynamics (C)
The work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of \(1 \mathrm{~L}\) to \(10 \mathrm{~L}\) at \(300 \mathrm{~K}\) is \((R=0.0083 \mathrm{~kJ} \mathrm{~K} \mathrm{~mol}=1)\)
- A \(5.8 \mathrm{~kJ}\)
- B \(0.115 \mathrm{~kJ}\)
- C \(58.5 \mathrm{~kJ}\)
- D \(11.5 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(D) \(11.5 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
Given, initial volume \(V_1=1 \mathrm{~L}\)
Final volume, \(V_2=2 \mathrm{~L}\)
Temperature \(=300 \mathrm{~K}\)
\[
R=0.0083 \mathrm{~kJ} \mathrm{~K} \mathrm{~mol}^{-1}
\]
\(\operatorname{Moles}(n)=2\) moles
We know that,
Work done, \(W=2.303 n R T \log \frac{V_2}{V_1}\)
\[
=2.303 \times 2 \times 0.0083 \times 300 \log 10=11.5 \mathrm{~kJ}
\]
Final volume, \(V_2=2 \mathrm{~L}\)
Temperature \(=300 \mathrm{~K}\)
\[
R=0.0083 \mathrm{~kJ} \mathrm{~K} \mathrm{~mol}^{-1}
\]
\(\operatorname{Moles}(n)=2\) moles
We know that,
Work done, \(W=2.303 n R T \log \frac{V_2}{V_1}\)
\[
=2.303 \times 2 \times 0.0083 \times 300 \log 10=11.5 \mathrm{~kJ}
\]
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