MHT CET · Chemistry · Thermodynamics (C)
An ideal gas absorbs \(210 \mathrm{~J}\) of heat and undergoes expansion from \(3 \mathrm{~L}\) to \(6 \mathrm{~L}\) against a constant external pressure of \(10^5 \mathrm{~Pa}\). What is the value of \(\Delta \mathrm{U}\) ?
- A \(310 \mathrm{~J}\)
- B \(-90 \mathrm{~J}\)
- C \(-210 \mathrm{~J}\)
- D \(190 \mathrm{~J}\)
Answer & Solution
Correct Answer
(B) \(-90 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\(1 \mathrm{~Pa}=1 \times 10^{-5}\) bar
\(\therefore \quad 10^5 \mathrm{~Pa}=1 \mathrm{bar}\)
\(\mathrm{W}=-\mathrm{P}_{\mathrm{ext}} \times \Delta \mathrm{V}=-1 \times(6-3)=-3 \mathrm{dm}^3\) bar
\(1 \mathrm{dm}^3 \mathrm{bar}=100 \mathrm{~J}\)
\(\therefore \quad \mathrm{W}=-300 \mathrm{~J}\)
\(\mathrm{Q}=+210 \mathrm{~J}\)
According to first law of thermodynamics,
\(\Delta \mathrm{U}=\mathrm{Q}+\mathrm{W}\)
\(\therefore \quad \Delta U=+210 \mathrm{~J}+(-300 \mathrm{~J})=-90 \mathrm{~J}\)
\(\therefore \quad 10^5 \mathrm{~Pa}=1 \mathrm{bar}\)
\(\mathrm{W}=-\mathrm{P}_{\mathrm{ext}} \times \Delta \mathrm{V}=-1 \times(6-3)=-3 \mathrm{dm}^3\) bar
\(1 \mathrm{dm}^3 \mathrm{bar}=100 \mathrm{~J}\)
\(\therefore \quad \mathrm{W}=-300 \mathrm{~J}\)
\(\mathrm{Q}=+210 \mathrm{~J}\)
According to first law of thermodynamics,
\(\Delta \mathrm{U}=\mathrm{Q}+\mathrm{W}\)
\(\therefore \quad \Delta U=+210 \mathrm{~J}+(-300 \mathrm{~J})=-90 \mathrm{~J}\)
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