Answer
a) $\rm 98\;cm^3$
b) See the graph below.
Work Step by Step
a) Since the gas undergoes an isobaric process, its pressure remains constant.
And we know, for an ideal gas, that
$$\dfrac{ \color{red}{\bf\not} P_1V_1}{T_1}=\dfrac{ \color{red}{\bf\not} P_2V_2}{T_2}$$
in our case, the pressure is constant.
$$\dfrac{ V_1}{T_1}=\dfrac{ V_2}{T_2}$$
We need to find the final volume of the gas, so we need to solve for $V_2$
$$V_2=\dfrac{ V_1T_2}{T_1}\tag 1$$
Plugging the known;
$$V_2=\dfrac{ (50\times 10^{-6})(300+273)}{(20+273)} =97.78\times 10^{-6}\;\rm m^3$$
$$V_1\approx \color{red}{\bf 98}\;\rm cm^3$$
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b) To draw a proper scale graph, we need to find the magnitude of the initial pressure (the constant pressure) which is given by
$$P_1V_1=nRT_1$$
Hence,
$$P_1=\dfrac{nRT_1}{V_1}$$
Plugging the known;
$$P=\dfrac{(0.1)(8.31)(20+273)}{(50\times 10^{-6})}=\bf 4.87\times 10^6\;\rm Pa$$
See the figure below.
