An ideal gas is taken from state A (Pressure P, Volume V) to the state B (Pressure P/2, Volume 2V) along a straight line path in PV diagram as shown in the adjacent figure.
Select the correct statement(s) among the following

Question

An ideal gas is taken from state A (Pressure P, Volume V) to the state B (Pressure P/2, Volume 2V) along a straight line path in PV diagram as shown in the adjacent figure.

Select the correct statement(s) among the following

TestHub · Accepted Answer

Statement A is correct because the straight-line path lies entirely above the convex-downward isothermal curve ( $P V = co n s t an t$ ), meaning the area under the line (work done) is greater. The path equation is a straight line with a negative slope: $P = α - β V$ (where $α, β$ are positive constants).

Using $P V = n RT$ , substituting $P$ gives $n RT = α V - β V^{2}$ , which represents a downward-opening parabola in the $T - V$ plane (Statement B is correct).

Temperature behavior: Differentiating $T$ with respect to $V$ shows that $T$ reaches a maximum value and then decreases as volume increases from $V$ to $2 V$ (Statement D is correct).

Statement C is incorrect because substituting $V = \frac{α - P}{β}$ into $P V = n RT$ yields $n RT = \frac{α P - P ^{2}}{β}$ , which is a parabola in the $P - T$ plane, not a hyperbola.

Chemistry - Thermodynamics - I Question with Solution | TestHub

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