Heat and Thermodynamics 4 Question 3
3. One mole of an ideal gas passes through a process, where pressure and volume obey the relation $p=p _0\left[1-\frac{1}{2}\left(\frac{V _0}{V}\right)^{2}\right]$.
Here, $p _0$ and $V _0$ are constants. Calculate the change in the temperature of the gas, if its volume changes from $V _0$ to $2 V _0$.
(a) $\frac{1}{2} \frac{p _0 V _0}{R}$
(b) $\frac{1}{4} \frac{p _0 V _0}{R}$
(c) $\frac{3}{4} \frac{p _0 V _0}{R}$
(d) $\frac{5}{4} \frac{p _0 V _0}{R}$
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Solution:
- Given process equation for 1 mole of an ideal gas is
$$ p=p _0\left(1-\frac{1}{2}\left(\frac{V _0}{V}\right)^{2}\right) $$
Also, for 1 mole of ideal gas,
$$ \begin{aligned} & p V=R T \\ \therefore \quad & p=\frac{R T}{V} \end{aligned} $$
So, from Eqs. (i) and (ii), we have
$$ \begin{aligned} & \frac{R T}{V} & =p _0\left(1-\frac{1}{2}\left(\frac{V _0}{V}\right)^{2}\right) \\ \therefore \quad & T & =\frac{p _0 V}{R}\left(1-\frac{1}{2}\left(\frac{V _0}{V}\right)^{2}\right) \end{aligned} $$
When volume of gas is $V _0$, then by substituting $V=V _0$ in Eq. (iii), we get
Temperature of gas is
$$ T _1=\frac{p _0 V _0}{R}\left(1-\frac{1}{2}\left(\frac{V _0}{V _0}\right)^{2}\right)=\frac{p _0 V _0}{2 R} $$
Similarly, at volume, $V=2 V _0$
Temperature of gas is
$$ T _2=\frac{p _0\left(2 V _0\right)}{R}\left(1-\frac{1}{2}\left(\frac{V _0}{2 V _0}\right)^{2}\right)=\frac{7}{4} \frac{p _0 V _0}{R} $$
So, change in temperature as volume changes from $V _0$ to $2 V _0$ is
$$ \Delta T=T _2-T _1=\left(\frac{7}{4}-\frac{1}{2}\right) \frac{p _0 V _0}{R}=\frac{5}{4} \frac{p _0 V _0}{R} $$
