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:

  1. 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

$$ \Delta T=T _2-T _1=\left(\frac{7}{4}-\frac{1}{2}\right) \frac{p _0 V _0}{R}=\frac{5}{4} \frac{p _0 V _0}{R} $$



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