91Ó°ÊÓ

Cylinder length$=20cm$

Temperature$={211}^{∘}C$

Given Data:

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

$p_1{V}_1^{\mathrm{γ}}=p_2{V}_2^{\mathrm{γ}}$

This allows us to express the ratio of pressures as

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

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

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

Having already expressed the ratio of pressures, we can write

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

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

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

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 $\mathrm{γ}-1$parameter. A such graph is given below:

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

$\mathrm{γ}-1=0.2918$

$\mathrm{γ}=1.292$