Answer
See the detailed answer below.
Work Step by Step
a) The process at which the pressure is directly proportional to the volume of the gas
$$P\propto V\tag 1$$
is a straight line with a positive slope since both increase or decrease at the same rate.
Now let's assume that the initial situation is $P_i$, $V_i$, $T_i$, and $v_{\rm rms,1}$.
Now we need to find the final situation at which the rms speed is doubled.
We know that the rms speed is given by
$$v_{\rm rms,1}=\sqrt{\dfrac{3k_BT_i}{m}}$$
And we know that the final rms speed is
$$v_{\rm rms,2}=2v_{\rm rms,1}=2\sqrt{\dfrac{3k_BT_i}{m}}$$
Hence,
$$v_{\rm rms,2} =2\sqrt{\dfrac{3k_BT_i}{m}}$$
$$\sqrt{\dfrac{3k_BT_f}{m}} =2\sqrt{\dfrac{3k_BT_i}{m}}$$
Squaring both sides
$$ \dfrac{ \color{red}{\bf\not} 3 \color{red}{\bf\not} k_BT_f}{ \color{red}{\bf\not} m} =(4)\dfrac{ \color{red}{\bf\not} 3 \color{red}{\bf\not} k_BT_i}{ \color{red}{\bf\not} m}$$
Thus,
$$T_f=4T_i\tag 2$$
Applying the ideal gas law,
$$\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}$$
Plugging from (2)
$$\dfrac{P_iV_i}{ \color{red}{\bf\not} T_i}=\dfrac{P_fV_f}{4 \color{red}{\bf\not} T_i}$$
Hence,
$$P_fV_f=4P_iV_i\tag 3$$
From (1), we can see that
$$\dfrac{P_i }{P_f}=\dfrac{V_i}{V_f}$$
Solving for $V_i$l
$$V_i=\dfrac{P_iV_f}{P_f}$$
Plugging into (3);
$$P_f \color{red}{\bf\not} V_f=4P_i \dfrac{P_i \color{red}{\bf\not} V_f}{P_f}$$
$$P_f^2=4P_i^2$$
Hence,
$$\boxed{P_f=2P_i}$$
And then the final volume is
$$\boxed{V_f=2V_i}$$
This means that the pressure is doubled and the volume is also doubled.
See the graph below.
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b) We know that the change in the thermal energy of the gas is given by
$$\Delta E_{th}=Q+W$$
Thus,
$$Q=\Delta E_{th}-W$$
where $W=-$Area under the $P-V$ curve, and $\Delta E_{th}=nC_{\rm V}\Delta T$
Hence,
$$Q=nC_{\rm V}(T_f-T_i)-{\rm(-Area)}$$
We are given that $C_{\rm V}=\frac{5}{2}R$
$$Q=\frac{5}{2}nR(T_f-T_i)+{\rm(Area)}$$
The area under the curve consists of two things, a right triangle and rectanguale, see the second figure below.
$$Q=\frac{5}{2}nR(4T_i-T_i)+\frac{1}{2}(2V_i-V_i)(2P_i-P_i)+P_i(2V_i-V_i)$$
$$Q=\frac{15}{2}nRT_i+ \frac{3}{2}P_iV_i\tag 4$$
Noting that $T_i$ is given by the ideal gas law
$$P_iV_i=nRT_i$$
Hence,
$$T_i=\dfrac{P_iV_i}{nR}$$
Plugging into (4);
$$Q=\frac{15}{2} \color{red}{\bf\not} n \color{red}{\bf\not} R \dfrac{P_iV_i}{ \color{red}{\bf\not} n \color{red}{\bf\not} R}+ \frac{3}{2}P_iV_i $$
$$Q=\frac{15}{2} P_iV_i + \frac{3}{2}P_iV_i $$
$$\boxed{Q=9P_iV_i}$$
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