Your laboratory assignment for the week is to measure the specific heat ratio $\gamma$of carbon dioxide. The gas is contained in a cylinder with a movable piston and a thermometer. When the piston is withdrawn as far as possible, the cylinder's length is $20\mathrm{cm}$. You decide to push the piston in very rapidly by various amounts and, for each push, to measure the temperature of the carbon dioxide. Before each push, you withdraw the piston all the way and wait several minutes for the gas to come to the room temperature of $211°\mathrm{C}$. Your data are as follows:

Use the best-fit line of an appropriate graph to determine $\gamma$for carbon dioxide.

Cylinder length$=20cm$

Temperature$={211}^{\circ }C$

Given Data:

Since the process is adiabatic, we know that $p{V}^{\gamma }=$const. Therefore we can write

$p_1{V}_1^{\gamma }=p_2{V}_2^{\gamma }$

This allows us to express the ratio of pressures as

$\frac{p_2}{p_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma }$

Rearranging the ideal gas law for the case when the number of moles is constant, we get

$\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}\Rightarrow \frac{p_2}{p_1}=\frac{V_1}{V_2}\frac{T_2}{T_1}$

Having already expressed the ratio of pressures, we can write

${\left(\frac{V_1}{V_2}\right)}^{\gamma }=\frac{V_1}{V_2}\frac{T_2}{T_1}$

Dividing both sides by the ratio of volumes and by the power rules, we can write

${\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=\frac{T_2}{T_1}$

Having done this derivation, what we can do is find out the ratios for all the compression trials. Also, we must take note that the ratio of the volume will be equal to the ratio of the initial (total) length of the cylinder by the length it reaches. That is, our list of volumes, in units of $V_1$, will be

$1.333,2.000,2.857,4.000\text{.}$

By a similar manner, we can construct a list of temperatures in the experiments, giving the temperatures after the compression in units of $T_1$. Considering that we must convert to Kelvin, we would have

$1.0476,1.1598,1.3027,1.4388$

Now we can plot the temperature ratios as a function of the volume ratios. After scattering, we can fit a power function and from that determine the $\gamma -1$parameter. A such graph is given below:

Therefore, as we can see from the fitting parameters, we have

$\gamma -1=0.2918$

$\gamma =1.292$