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 \( 9 P_1 V_1 \).

The root mean square (r.m.s.) speed \( v_{\text{rms}} \) for \( n \) molecules with speeds \( v_1, v_2, \ldots, v_n \) is calculated by:

\[
v_{\text{rms}} = \sqrt{\frac{v_1^2 + v_2^2 + v_3^2 + \cdots + v_n^2}{n}}
\]

Given speeds are: \( 2, 3, 4, 5, 6 \).

1. Calculate the squares:
\[
2^2 = 4, \quad 3^2 = 9, \quad 4^2 = 16, \quad 5^2 = 25, \quad 6^2 = 36
\]

2. Sum of the squares:
\[
4 + 9 + 16 + 25 + 36 = 90
\]

3. Divide by the number of molecules (5) and take the square root:
\[
v_{\text{rms}} = \sqrt{\frac{90}{5}} = \sqrt{18} \approx 4.24
\]

Thus, the r.m.s. speed is approximately \( 4.24 \).

The root mean square (r.m.s.) speed \( v_{\text{rms}} \) of gas molecules is given by the formula:

\[
v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}
\]

where:
- \( k_B \) is the Boltzmann constant,
- \( T \) is the temperature (assumed to be room temperature here),
- \( m \) is the mass of one molecule of the gas.

Rearranging to find \( m \):

\[
m = \frac{3 k_B T}{v_{\text{rms}}^2}
\]

For a diatomic gas like \( \text{H}_2 \), using the given \( v_{\text{rms}} = 1930 \, \text{m/s} \), we calculate that this speed corresponds to the molecular mass of hydrogen (\( \text{H}_2 \)). Therefore, the gas is identified as \( \text{H}_2 \).

For a mixture of \( n_1 \) moles of a monatomic gas and \( n_2 \) moles of a diatomic gas, the ratio \( \gamma = \frac{C_p}{C_v} \) of the mixture is given by:

\[
\gamma = \frac{\text{Total } C_p}{\text{Total } C_v}
\]

1. Molar heat capacities:
- For a monatomic gas: \( C_{v, \text{mono}} = \frac{3}{2} R \) and \( C_{p, \text{mono}} = \frac{5}{2} R \).
- For a diatomic gas: \( C_{v, \text{di}} = \frac{5}{2} R \) and \( C_{p, \text{di}} = \frac{7}{2} R \).

2. Total heat capacities:
- Total \( C_v = n_1 \cdot \frac{3}{2} R + n_2 \cdot \frac{5}{2} R \).
- Total \( C_p = n_1 \cdot \frac{5}{2} R + n_2 \cdot \frac{7}{2} R \).

3. Given \( \gamma = 1.5 \):

\[
\frac{C_p}{C_v} = \frac{n_1 \cdot \frac{5}{2} R + n_2 \cdot \frac{7}{2} R}{n_1 \cdot \frac{3}{2} R + n_2 \cdot \frac{5}{2} R} = 1.5
\]

4. Simplify by canceling \( R \) and multiplying through by 2:

\[
\frac{5n_1 + 7n_2}{3n_1 + 5n_2} = 1.5
\]

5. Cross-multiply to solve for \( n_1 \) in terms of \( n_2 \):

\[
5n_1 + 7n_2 = 1.5 (3n_1 + 5n_2)
\]

\[
5n_1 + 7n_2 = 4.5n_1 + 7.5n_2
\]

6. Rearrange terms:

\[
0.5n_1 = 0.5n_2
\]

\[
n_1 = n_2
\]

To have the same rms speed for \(\text{H}_2\) and \(\text{N}_2\), we use the formula for rms speed:

\[
v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}
\]

Since the rms speeds are equal, we can set up the equation:

\[
\sqrt{\frac{3k_B T_{\text{H}_2}}{m_{\text{H}_2}}} = \sqrt{\frac{3k_B T_{\text{N}_2}}{m_{\text{N}_2}}}
\]

Square both sides and simplify:

\[
\frac{T_{\text{H}_2}}{m_{\text{H}_2}} = \frac{T_{\text{N}_2}}{m_{\text{N}_2}}
\]

Since \(\text{N}_2\) is 14 times heavier than \(\text{H}_2\), we have \( m_{\text{N}_2} = 14 \, m_{\text{H}_2} \) and \( T_{\text{N}_2} = 27^\circ \text{C} = 300 \, \text{K} \).

Now solve for \( T_{\text{H}_2} \):

\[
T_{\text{H}_2} = \frac{T_{\text{N}_2}}{14} = \frac{300}{14} \approx 21.4 \, \text{K}
\]

So, the temperature at which \(\text{H}_2\) has the same rms speed as \(\text{N}_2\) at 27°C is approximately \( 21.4 \, \text{K} \).

The root mean square (rms) speed \( v_{\text{rms}} \) of a gas is given by:

\[
v_{\text{rms}} = \sqrt{\frac{3RT}{M}}
\]

where \( M \) is the molar mass.

For helium (\( M = 4 \, \text{g/mol} \)) and argon (\( M = 40 \, \text{g/mol} \)) at the same temperature, the ratio of their rms speeds is:

\[
\frac{v_{\text{rms}(\text{He})}}{v_{\text{rms}(\text{Ar})}} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16
\]

Thus, the ratio \( \frac{v_{\text{rms}(\text{He})}}{v_{\text{rms}(\text{Ar})}} \) is approximately 3.16.

For a gas mixture of oxygen (\(\text{O}_2\)) and argon (\(\text{Ar}\)):

1. **Oxygen (\(\text{O}_2\))** is diatomic, so its internal energy per mole is:
\[
U_{\text{O}_2} = \frac{5}{2} RT
\]
For 2 moles, \( U_{\text{O}_2} = 2 \times \frac{5}{2} RT = 5 RT \).

2. **Argon (\(\text{Ar}\))** is monatomic, so its internal energy per mole is:
\[
U_{\text{Ar}} = \frac{3}{2} RT
\]
For 4 moles, \( U_{\text{Ar}} = 4 \times \frac{3}{2} RT = 6 RT \).

3. **Total internal energy**:
\[
U_{\text{total}} = U_{\text{O}_2} + U_{\text{Ar}} = 5 RT + 6 RT = 11 RT
\]

So, the total internal energy of the system is \( 11 RT \).

The mean kinetic energy of gas molecules depends only on temperature and is given by:

\[
\text{KE} = \frac{3}{2} k_B T
\]

where \( T \) is the temperature in Kelvin, and \( k_B \) is Boltzmann's constant.

Since we want the mean kinetic energy of \(\text{O}_2\) to equal that of \(\text{H}_2\) at \( -73^\circ \text{C} \):

1. Convert \(-73^\circ \text{C}\) to Kelvin:
\[
T_{\text{H}_2} = -73 + 273 = 200 \, \text{K}
\]

2. Since kinetic energy depends only on temperature, set the temperature \( T_{\text{O}_2} = T_{\text{H}_2} = 200 \, \text{K} \).

Therefore, the temperature at which \(\text{O}_2\) has the same mean kinetic energy as \(\text{H}_2\) at \(-73^\circ \text{C}\) is \(200 \, \text{K}\).

For an ideal gas obeying \( PV^{\frac{1}{2}} = \text{constant} \):

1. Use the ideal gas law: \( PV = nRT \), so \( P = \frac{nRT}{V} \).
2. Substitute \( P = \frac{nRT}{V} \) in \( PV^{\frac{1}{2}} = \text{constant} \):

\[
\frac{nRT}{V} \cdot V^{\frac{1}{2}} = \text{constant} \Rightarrow nRT \cdot V^{-\frac{1}{2}} = \text{constant}
\]

3. At initial state, \( T = T \) and \( V = V \):

\[
T V^{-\frac{1}{2}} = \text{constant}
\]

4. When \( V \) changes to \( 4V \):

\[
T' (4V)^{-\frac{1}{2}} = T V^{-\frac{1}{2}}
\]

5. Simplify:

\[
T' \cdot \frac{1}{2} = T \Rightarrow T' = 2T
\]

So, the final temperature \( T' = 2T \).