Solution:

Since the pressure \( P \) is directly proportional to the volume \( V \), we have:

\[
P \propto V \Rightarrow P = kV
\]

where \( k \) is a constant. This process describes a line through the origin on a \( P \)-\( V \) graph, where work \( W \) done by the gas is given by the area under the line:

\[
W = \frac{P_{\text{final}} V_{\text{final}} - P_{\text{initial}} V_{\text{initial}}}{2}
\]

Given that the r.m.s. speed doubles, we know the temperature \( T \) has quadrupled (since \( v_{\text{rms}} \propto \sqrt{T} \)). Therefore:

\[
\frac{T_{\text{final}}}{T_{\text{initial}}} = 4
\]

For an ideal gas, \( PV = nRT \), so if \( T \) quadruples, \( PV \) also quadruples, making \( P_{\text{final}} V_{\text{final}} = 4 P_1 V_1 \).

Using the work formula above:

\[
W = \frac{4 P_1 V_1 - P_1 V_1}{2} = \frac{3 P_1 V_1}{2} = 3 P_1 V_1
\]

The change in internal energy \( \Delta U \) for a diatomic gas (\( C_V = \frac{5}{2}R \)) is:

\[
\Delta U = n C_V \Delta T = \frac{5}{2} (P_1 V_1) \cdot 3 = \frac{15}{2} P_1 V_1 = 7.5 P_1 V_1
\]

Using the first law of thermodynamics \( Q = \Delta U + W \):

\[
Q = 7.5 P_1 V_1 + 3 P_1 V_1 = 9 P_1 V_1
\]

Answer: The heat supplied to the gas is \( 9 P_1 V_1 \).