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An ideal gas is taken from state A (pressure P, volume V) to state B (pressure P/2, volume 2V) along a straight line in the P − V diagram as shown in fig. Then which is incorrect

Option: 1

The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along the isotherm


Option: 2

In the T-V diagram, the path AB becomes a part of a parabola


Option: 3

In the P-T diagram, the path AB becomes a part of a hyperbola


Option: 4

In going from A to B, the temperature T of the gas first increases to a maximum and then decreases


Answers (1)

best_answer

(a) Work done in process A to B is (see fig.)
\mathrm{W_1} = area of trapezium ABCD

\mathrm{\begin{aligned} & =\left(P+\frac{P}{2}\right) V=\frac{3 P V}{2} \\ & =\frac{3 R T}{2}(\text { for } 1 \text { mole }) \end{aligned}}

If the process A to B were isothermal, the work done would be

\mathrm{W_2=R T \log _e\left(\frac{V_2}{V_1}\right)=R T \log _e(2)=0.69 R T}

\mathrm{\text { Thus } W_1>W_2 \text {. So choice (a) is correct }}

\mathrm{\text { (b) Let } P_0 \text { and } V_0 \text { be the intercepts on the } P \text { and } V \text { axes. }}

The equation of straight line AB is

\mathrm{\begin{aligned} & P=-\frac{P_0}{V_0}\left(V-V_0\right) \\ & \Rightarrow \frac{P}{P_0}+\frac{V}{V_0}=1 \end{aligned}}   ....[1]

\mathrm{\text { Since } P=\frac{R T}{V} \text {, Eq. (1) becomes }}

\mathrm{\frac{R T}{V P_0}+\frac{V}{V_0}=1 \Rightarrow T=\frac{P_0 V}{R}-\frac{P_0 V^2}{R V_0}}

Which represents a parabola on the T-V graph
So choice (b) is also correct

\mathrm{\text { (c) Since } V=\frac{R T}{P} \text {, Eq. (1) becomes }}

\mathrm{\frac{P}{P_0}+\frac{R T}{P V_0}=1 \Rightarrow T=V_0 P-\frac{V_0 P^2}{R P_0}}       ......[2]

Which does not represent a hyperbola. So choice (c) is incorrect
(d) If follows from Eq. (2) above that choice (d) is correct

 

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Anam Khan

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