Heat and Thermodynamics 6 Question 21
26. An ideal gas is taken from the state $A$ (pressure $p$, volume $V$ ) to the state $B$ (pressure $p / 2$, volume $2 V$ ) along a straight line path in the $p$ - $V$ diagram. Select the correct statements from the following
$(1993,2 M)$
(a) 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 an isotherm
(b) In the $T-V$ diagram, the path $A B$ becomes a part of a parabola
(c) In the $p-T$ diagram, the path $A B$ becomes a part of a hyperbola
(d) In going from $A$ to $B$, the temperature $T$ of the gas first increases to a maximum value and then decreases
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Answer:
Correct Answer: 26. (a, b, d)
Solution:
- (a) Work done $=$ Area under $p-V$ graph
$$ \begin{aligned} A _1 & >A _2 \\ \therefore \quad W _{\text {given process }} & >W _{\text {isothermal process }} \end{aligned} $$
(b) In the given process $p$ - $V$ equation will be of a straight line with negative slope and positive intercept i.e., $p=-\alpha V+\beta$ (Here $\alpha$ and $\beta$ are positive constants)
$\Rightarrow \quad p V=-\alpha V^{2}+\beta V$
$$ \begin{array}{ll} \Rightarrow & n R T=-\alpha V^{2}+\beta V \\ \Rightarrow & T=\frac{1}{n R}\left(-\alpha V^{2}+\beta V\right) \cdots(i) \end{array} $$
This is an equation of parabola in $T$ and $V$.
(d) $\frac{d T}{d V}=0=\beta-2 \alpha V$
$$ \begin{array}{ll} \Rightarrow & V=\frac{B}{2 \alpha} \\ \text { Now, } & \frac{d^{2} T}{d V^{2}}=-2 \alpha=-ve \end{array} $$
ie, $T$ has some maximum value.
$$ \begin{array}{rlrl} & \text { Now, } & T & \propto p V \\ \text { and } & & (p A) _A & =(p V) _B \\ \Rightarrow & T _A & =T _B \end{array} $$
We conclude that temperatures are same at $A$ and $B$ and in between temperature has a maximum value. Therefore, in going from $A$ to $B, T$ will first increase to a maximum value and then decrease.
