Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IWorkMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

Two moles of an ideal gas $(γ = 5/3)$ are initially at a temperature of 27°C and occupy a volume of 20 litre. The gas is first expanded at constant pressure until the volume is doubled. It then undergoes reversible adiabatic change until the temperature returns to its initial value. ( $R = 0.082 L - atm / K - mol$ )

The total work done (in Kcal) by the gas is.

Answer:

3.00

Solution:

Isobaric Work ( $W_{1}$ ): Doubling the volume at constant pressure doubles the temperature to $600 K$ , so $W_{1} = n R Δ T = 2 \times R \times (600 - 300) = 600 R$ .

Adiabatic Work ( $W_{2}$ ): Cooling back to $300 K$ yields $W_{2} = \frac{n R ( T _{1} - T _{2} )}{γ - 1} = \frac{2 \times R \times ( 600 - 300 )}{( 5/3 ) - 1} = 900 R$ .

Total Work ( $W$ ): Summing both steps gives $W = W_{1} + W_{2} = 600 R + 900 R = 1500 R$ .

Final Conversion: Using $R \approx 2 c a l / m o l . k$ gives $1500 \times 2 = 3000 c a l = 3.00Kcal$ .

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Work

⏱ 2mℹ️ Source: QB

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