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;
$$n=\dfrac{14\;\rm g}{\rm 2(14)\;g/mol}=\bf 0.5\;\rm mol $$
Noting that 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,
$$V1=\dfrac{nRT_1}{P_1}=\dfrac{(0.5)(8.31)(0+273)}{(1.013×10^5)}$$
$$V_1=\bf 11.2\times 10^{-3}\;\rm m^3$$
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a) For an isochoric process, the final temperature is given by
$$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$$
Hence,
$$T_2=\dfrac{P_2T_1}{P_1}$$
Plugging the known;
$$T_2=\dfrac{(20)(0+273)}{1}$$
$$T_2=\color{red}{\bf 5460}\;\rm K$$
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b) For an isochoric process, the work done on the gas is zero since the volume is constant.
$$W=\color{red}{\bf 0}\;\rm J$$
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c) We know that the change in thermal energy of the gas during any process is given by
$$\Delta E_{th}=Q+W=Q+0$$
Hence,
$$Q=\Delta E_{th}=nC_{\rm v}(T_2-T_1)$$
Plugging the known;
$$Q= (0.5)(20.8)(5460−273)$$
$$Q=\color{red}{\bf 5.39\times 10^4}\;\rm J$$
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d) For an isochoric process,
$$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}=\dfrac{5460}{273}=\bf 20 $$
This means that $P_2=20P_1$ which means that $P_2$ is the maximum pressure.
$$\dfrac{P_{\rm max}}{P_{\rm min}}= \color{red}{\bf 20 }$$
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e) To draw this process, we need to know the initial and final points.
The initial point is: $(V_1,P_1)=\rm ( 11,200\;\rm cm^3,1\;atm)$
And the final point is: $(V_2,P_2)=\rm ( 11,200\;\rm cm^3,20\;atm)$
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