Answer
See the detailed answer below.
Work Step by Step
First of all, we need to find the number of moles of this gas
$$n=\dfrac{m}{M_{N_2}}=\dfrac{m}{2M_N}$$
Plugging the known;
$$n =\dfrac{14\;\rm g}{2(14)\;\rm g/mol}=\bf 0.5\;\rm mol$$
Noting that $\rm STP$ means standard temperature and pressure at which the temperature is 0$^\circ$C and the pressure is 1 atm.
So we can find the initial volume by applying the ideal gas law;
$$PV=nRT$$
Hence,
$$V_1=\dfrac{nRT_1}{P_1}=\dfrac{(0.5)(8.31)(0+273)}{(1.013\times 10^5)}$$
$$V_1=\bf 11.2\times 10^{-3}\;\rm m^3$$
a) For an adiabatic process, the final temperature is given by
$$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$
Hence,
$$\dfrac{T_2}{T_1}=\left[\dfrac{V_1}{V_2} \right]^{\gamma-1}$$
$$T_2=T_1\left[\dfrac{V_1}{V_2} \right]^{\gamma-1}\tag 1$$
Now we need to find the ratio of volumes in terms of pressure.
$$P_1V_1^\gamma=P_2V_2^\gamma$$
Hence,
$$\left[\dfrac{V_1}{V_2}\right]^\gamma= \dfrac{P_2}{P_1} $$
$$\dfrac{V_1}{V_2}=\left[ \dfrac{P_2}{P_1}\right]^{\frac{1}{\gamma}}\tag 2 $$
Plugging into (1);
$$T_2=T_1\left[ \left[ \dfrac{P_2}{P_1}\right]^{\frac{1}{\gamma}} \right]^{\gamma-1} $$
$$T_2=T_1\left[ \dfrac{P_2}{P_1} \right] ^{\frac{\gamma-1}{\gamma}} $$
Plugging the known;
$$T_2=(0+273)\left[ \dfrac{20}{1} \right] ^{\frac{1.4-1}{1.4}} $$
$$T_2=\color{red} {\bf 643}\;\rm K=\color{red} {\bf 370}^\circ \rm C$$
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b) We know that the change in thermal energy of the gas during any process is given by
$$\Delta E_{th}=Q+W$$
And we know that the heat exchange $Q$ during an adiabatic process is zero.
So, the work done on the gas is given by:
$$W=\Delta E_{th}-Q=\Delta E_{th}-0$$
$$W=nC_{\rm v}(T_2-T_1)$$
Plugging the known;
$$W=(0.5) (20.8) (643-273)$$
$$W=\color{red}{\bf 3.85\times 10^3}\;\rm J$$
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c) As we mentioned above, in part b,
$$Q=\color{red}{\bf 0}\;\rm J$$
during any adiabatic process.
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d) We need to find the ratio in (2);
$$\dfrac{V_1}{V_2}=\left[ \dfrac{P_2}{P_1}\right]^{\frac{1}{\gamma}} =\left[ \dfrac{20}{1}\right]^{\frac{1}{1.4}}=\bf 8.5 \tag 3 $$
This means that the initial volume is about 9 times the final volume. Thus, the initial volume is the maximum volume.
Thus
$$\dfrac{V_{\rm max}}{V_{\rm min}}= \color{red}{\bf 8.5 } $$
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e) To draw this process, we need to find the two volumes in L.
$$V_1=11.2\times 10^{-3}\times 10^3=\bf 11.2\;\rm L$$
$V_2$ is given from (3)
$$V_2=\dfrac{V_1}{8.5}=\dfrac{11.2}{8.5}=\bf 1.32\;\rm L$$
Hence,
The initial point is: $(V_1,P_1)=\rm (11.200\;L,1\;atm)$
And the final point is: $(V_2,P_2)=\rm (1.318\;L,20\;atm)$
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