The final volume (in L) of gas is. | TestHub

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 final volume (in L) of gas is.

Answer:

113.12

Solution:

Step 1: In the isobaric expansion ( $1 \to 2$ ), volume doubles to $V_{2} = 40 L$ , so temperature also doubles to $T_{2} = 2 \times 300 K = 600 K$ .

Step 2: In the adiabatic process ( $2 \to 3$ ), the temperature returns to its initial value, so $T_{3} = 300 K$ .

Step 3: Use the adiabatic relation $T_{2} V_{2}^{γ - 1} = T_{3} V_{3}^{γ - 1}$ , where $γ - 1 = \frac{5}{3} - 1 = \frac{2}{3}$ .

Step 4: Substitute the values: $600 \times (40)^{2/3} = 300 \times (V_{3})^{2/3}$ , which simplifies to $(V_{3})^{2/3} = 2 \times (40)^{2/3}$ .

Step 5: Solve for final volume: $V_{3} = 40 \times 2^{3/2} = 40 \times 2.8284 \approx 113.14L$ .

Chemistry - Thermodynamics - I Question with Solution | TestHub

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