A container of volume 2L is separated into equal compartments. In one compartment one mole of an ideal monoatomic gas is filled at 1 bar pressure and the other compartment is completely evacuated. A pinhole is made in the separator so gas expands to occupy full 2L and heat is supplied to gas so that finally pressure of gas equals 1 bar. The correct statement(s) is/are: (Given : 1 liter bar = 100J)

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$∆$ U = 150J

$∆$ H = 250J

Entropy change of system is greater than zero

Work done by gas is zero

answer is A, B, C, D.

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Detailed Solution

$\begin{array}{l} T_{i} = \frac{P_{i} V_{i}}{nR}; T_{i} = \frac{1 \times 1}{1 \times R} = \frac{1}{R}; \\ T_{f} = \frac{P_{f} V_{f}}{nR} \\ T_{f} = \frac{1 \times 2}{1 \times R} = \frac{2}{R}; T_{f} = 2 T_{i}; ΔU = {nC}_{V} ΔT = 1 \times \frac{3}{2} \times R \times (T_{f} - T_{i}) \\ ΔU = \frac{3}{2} \times R \times T_{i} = \frac{3}{2} \times P_{i} V_{i} \end{array}$
$ΔU = \frac{3}{2} \times 1 \times 1 = \frac{3}{2}$ litre bar = 3/2 x 100 Joule
$∆$ U = 150 Joule

H = U + PV
$ΔH = ΔU + Δ (PV) = \frac{3}{2} P_{i} V_{i} + (P_{f} V_{f} - P_{i} V_{i})$
$ΔH = \frac{3}{2} P_{i} V_{i} + P_{i} V_{i} ΔH = (\frac{3}{2} + 1)$ litre bar
$ΔH = \frac{5}{2} \times 100 J ΔH = 250 J$
Entropy change is positive
Work done by gas is zero

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