Question

During an experiment, an ideal gas is found to obey a condition $V{P}^2=constant$. The gas is initially at a temperature $T$, pressure $P$ and volume $V$. The gas expands to volume $4V$. Then,

• A. the pressure of gas changes to $\frac{P}{2}$
• B. the temperature of gas changes to $4T$
• C. the graph of above process on the $P-T$ diagram is parabola.
• D. the graph of above process on a $P-T$ diagram is hyperbola

Answer 1

Answer: (A), (D)

the pressure of gas changes to $\frac{P}{2}$
the graph of above process on a $P-T$ diagram is hyperbola

During an adiabatic process $P{V}^{\gamma }=constant$ ....(1)
Given,
$V{P}^2=constant$
$\therefore V_1{P}_1^2=V_2{P}_2^2=constant$
Given $V_2=4V_1$ $\Rightarrow V_1{P}_1^2=4V_1{P}_2^2$ $\Rightarrow P_2=\frac{P_1}{2}$
$\therefore P{V}^{0.5}=constant$ .....(2)
Comparing (2) with (1) we get, $\gamma =\frac{1}{2}$
Option (B)
For adiabatic process, $T{V}^{\gamma -1}=constant$
$\therefore T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
$\therefore \frac{T_2}{T_1}=(\frac{1}{4}{)}^{0.5-1}=(\frac{1}{4}{)}^{-0.5}=2^{(-2)\times (-0.5)}=2$
Option(C) and (D):
For adiabatic process, $\frac{T^{\gamma }}{P^{\gamma -1}}=constant$

Substituting $\gamma =\frac{1}{2}$, we get $\frac{T^{0.5}}{P^{0.5-1}}=constant$
$\therefore TP=constant$
$\therefore$ PT curve is not a parabola, but a hyperbola