Answer
See the detailed answer below.
Work Step by Step
First of all, we need to find the number of moles of nitrogen gas which is given by
$$n=\dfrac{m}{M_{N_2}}=\dfrac{m}{2M_N}$$
Plugging the known;
$$n=\dfrac{1}{2(14)}=\bf\frac{1}{28}\;\rm mol$$
$\star$ a)
$\bullet$ At point 1: the pressure is given by
$$P_1V_1=nRT_1$$
$$P_1=\dfrac{nRT_1}{V_1}$$
Plugging the known;
$$P_1=\dfrac{(\frac{1}{28})(8.31)(25+273)}{(100\times 10^{-6})}$$
$$P_1=\color{red}{\bf 8.84\times 10^5}\;\rm Pa$$
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$\star$ b)
$\bullet$ At point 2: the temperature in Celsius is given by
$$T_2=\dfrac{P_2V_2}{nR}-273$$
From the given graph, we can see that $P_2=2P_1$, $V_2=V_1=100\;\rm cm^3$.
Plugging the known;
$$T_2=\dfrac{(2\times 8.84\times 10^5)(100\times 10^{-6})}{ (\frac{1}{28}) (8.31)}-273$$
$$T_2=\color{red}{\bf 323}^\circ\rm C$$
$\bullet$ At point 3: the temperature in Celsius is given by
$$T_3=\dfrac{P_3V_3}{nR}-273$$
From the given graph, we can see that $P_3=1.5P_1$, $V_3=0.5V_1=50\;\rm cm^3$.
Plugging the known;
$$T_3=\dfrac{(1.5\times 8.84\times 10^5)(50\times 10^{-6})}{ (\frac{1}{28}) (8.31)}-273$$
$$T_3=\color{red}{\bf -49.6}^\circ\rm C$$
$\bullet$ At point 4: the temperature in Celsius is given by
$$T_4=\dfrac{P_4V_4}{nR}-273$$
From the given graph, we can see that $P_4=1.5P_1$, $V_4=150\;\rm cm^3$.
Plugging the known;
$$T_4=\dfrac{(1.5\times 8.84\times 10^5)(150\times 10^{-6})}{ (\frac{1}{28}) (8.31)}-273$$
$$T_4=\color{red}{\bf 397}^\circ\rm C$$
$\rightarrow$ You can use the ideal gas law of $$\dfrac{P_1V_1 }{T_1}=\dfrac{P_fV_f }{T_f}$$
and you will get the same answer for all stages.
