Answer
See the detailed answer below.
Work Step by Step
a) When the gas undergoes an isochoric process, then
$$Q=nC_{\rm v}\Delta T$$
$$Q=nC_{\rm v}(T_f-T_i)$$
Thus,
$$T_f=\dfrac{Q}{nC_{\rm v}}+T_i\tag 1$$
Now we need to find the number of moles of this 3 g-Helium
$$n=\dfrac{m}{M_{He}}=\dfrac{3}{4}=\bf 0.75\;\rm mol$$
Plugging into (1) and plugging the other known;
$$T_f=\dfrac{1000}{(0.75)(12.5) }+(20+273)\approx\bf 400\;\rm K$$
The final pressure is given by
$$\dfrac{P_i\color{red}{\bf\not} V_i}{T_i}=\dfrac{P_f\color{red}{\bf\not} V_f}{T_f}$$
$$P_f=\dfrac{P_iT_f}{T_i}\tag 2$$
Now we need to find $P_i$ which is given by
$$P_iV_i=nRT_i$$
Hence,
$$P_i=\dfrac{nRT_i}{V_i}$$
Plugging into (2);
$$P_f=\dfrac{nR\color{red}{\bf\not} T_i T_f}{V_i\color{red}{\bf\not} T_i} $$
$$P_f=\dfrac{nR T_f}{V_i } =\dfrac{nR T_f}{L^3 } $$
Plugging the known;
$$P_f=\dfrac{(0.75)(8.31)(400)}{(0.20^3 )} $$
$$P_f=3.12\times 10^5\;\rm Pa\approx \color{red}{\bf3.1}\;\rm atm$$
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b) When the gas undergoes an isobaric process, then
$$Q=nC_{\rm p}\Delta T$$
$$Q=nC_{\rm p}(T_f-T_i)$$
Thus,
$$T_f=\dfrac{Q}{nC_{\rm p}}+T_i $$
Plugging the known;
$$T_f=\dfrac{1000}{(0.75)(20.8)}+(20+273)=\bf 357\;\rm K$$
The final volume is given by
$$\dfrac{\color{red}{\bf\not}P_iV_i}{T_i}=\dfrac{\color{red}{\bf\not} P_fV_f}{T_f}$$
$$V_f=\dfrac{V_iT_f}{T_i} =\dfrac{L^3T_f}{T_i} $$
Plugging the known;
$$V_f =\dfrac{(20)^3(357)}{(20+273)} =\bf 9.75\times 10^3\;\rm cm^3$$
$$V_f=\color{red}{\bf 9.75}\;\rm L$$
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c) To draw the graph, we need to know the initial and final points of each process.
- For the isochoric process;
The initial point: $\rm (8\;\rm L, 2.25\;atm)$
The final point: $\rm (8\;\rm L, 3.1\;atm)$
- For the isobaric process;
The initial point: $\rm (8\;\rm L, 2.25\;atm)$
The final point: $\rm (9.75\;\rm L, 2.25\;atm)$
