Answer
$1.29$
Work Step by Step
According to the given information, the gas undergoes an adiabatic process each time you push the piston.
We know, for an ideal gas that undergoes an adiabatic process, that
$$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$
Hence,
$$\dfrac{T_1}{T_2}=\left[\dfrac{V_2}{V_1}\right]^{\gamma-1}$$
Taking the natural logarithm for both sides;
$$\ln\left[\dfrac{T_1}{T_2}\right]=\ln \left[\dfrac{V_2}{V_1}\right]^{\gamma-1}=(\gamma-1)\ln \left[\dfrac{V_2}{V_1}\right]$$
$$\ln\left[\dfrac{T_1}{T_2}\right]= (\gamma-1)\ln \left[\dfrac{V_2}{V_1}\right]$$
The volume of the cylinder is given by $V=Ah$ and $A$, which is the cross-sectional area, is constant.
$$\ln\left[\dfrac{T_1}{T_2}\right]=( \gamma-1)\ln \left[\dfrac{\color{red}{\bf\not} Ah_2}{\color{red}{\bf\not} Ah_1}\right] $$
$$\ln\left[\dfrac{T_1}{T_2}\right]=( \gamma-1)\ln \left[\dfrac{ h_2}{ h_1}\right] $$
Plugging the initial case 1;
$$\ln\left[\dfrac{21+273}{T_2}\right]=( \gamma-1)\ln \left[\dfrac{h_2}{20}\right] $$
neglect the subtitle 2;
$$\boxed{\ln\left[\dfrac{294}{T}\right]=( \gamma-1)\ln \left[\dfrac{h}{20}\right] }$$
This is a straight-line formula of $y=mx+a$ where $y=\ln\left[\dfrac{294}{T_2}\right]$, $x=\ln \left[\dfrac{h}{20}\right] $, and the slope $m= \gamma-1$
Noting that $h=20-{\rm pushed\;distance}$
Thus,
$$\gamma={\rm Slope}+1\tag 1$$
\begin{array}{|c|c|c|}
\hline
x=\ln \left[\dfrac{h}{20}\right] & y=\ln\left[\dfrac{294}{T}\right]\\
\hline
\ln \left[\dfrac{15}{20}\right]& \ln\left[\dfrac{294}{308}\right]\\
\hline
\ln \left[\dfrac{10}{20}\right]& \ln\left[\dfrac{294}{341}\right] \\
\hline
\ln \left[\dfrac{7}{20}\right]& \ln\left[\dfrac{294}{383}\right] \\
\hline
\ln \left[\dfrac{5}{20}\right]& \ln\left[\dfrac{294}{423}\right] \\
\hline
\end{array}
Using any software calculator, we got the following result of slope (0.29), as we see below.
Plugging into (1);
$$\gamma=0.29+1=\color{red}{\bf 1.29} $$
