Since the gases are mixed at constant volume \( V \) and temperature \( T \), we can apply the ideal gas law for each gas:

For each gas, we have:
\[
P = \frac{nRT}{V}
\]

Since both gases have the same pressure \( P \), volume \( V \), and temperature \( T \), they contribute equally to the total pressure when mixed.

After mixing, the total pressure of the mixture is the sum of the partial pressures of each gas:
\[
P_{\text{total}} = P + P = 2P
\]

Thus, the pressure of the mixture is 2P.

To solve this, we use the relationship from Charles's Law, which states that for a gas at constant pressure:

\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\]

where:
- \( V_1 \) is the initial volume at temperature \( T_1 \),
- \( V_2 \) is the final volume at temperature \( T_2 \).

Given:
- Initial temperature \( T_1 = 0^\circ \text{C} = 273 \, \text{K} \),
- \( V_2 = 2V_1 \) (double the initial volume).

Now, substituting into Charles's Law:

\[
\frac{V_1}{273} = \frac{2V_1}{T_2}
\]

Solving for \( T_2 \):

\[
T_2 = 2 \times 273 = 546 \, \text{K}
\]

Converting \( T_2 \) back to Celsius:

\[
T_2 = 546 - 273 = 273^\circ \text{C}
\]

So, the required temperature is 273°C.

Given that the gas obeys the law \( VP^2 = \text{constant} \).

1. Initially:
\[
VP^2 = k
\]

2. When the volume changes from \( V \) to \( 2V \):
\[
(2V)P'^2 = k
\]
where \( P' \) is the new pressure.

Since \( VP^2 = (2V)P'^2 \), we can relate the pressures as:
\[
P'^2 = \frac{P^2}{2}
\]
\[
P' = \frac{P}{\sqrt{2}}
\]

3. Use the ideal gas law initially and finally:
\[
PV = nRT
\]
\[
P' \cdot 2V = nRT'
\]

Substitute \( P' = \frac{P}{\sqrt{2}} \):
\[
\frac{P}{\sqrt{2}} \cdot 2V = nRT'
\]

4. Simplify:
\[
\sqrt{2} PV = nRT'
\]

Since \( PV = nRT \):
\[
\sqrt{2} \cdot nRT = nRT'
\]
\[
T' = \sqrt{2} T
\]

So, the new temperature is \( \sqrt{2} T \).

To determine which line corresponds to each type of gas (monoatomic, diatomic, polyatomic), we can use the fact that the specific heat at constant pressure \( C_p \) varies with the degrees of freedom of each gas. Since \( Q = n C_p \Delta T \), for a given \( Q \), the slope of the \( Q \)-\( \Delta T \) line is inversely proportional to \( C_p \).

1. Monoatomic gas (M): \( C_p = \frac{5}{2} R \).
2. Diatomic gas (D): \( C_p = \frac{7}{2} R \).
3. Polyatomic gas (P): \( C_p \) is higher than both monoatomic and diatomic due to additional rotational degrees of freedom.

Since the slope is inversely related to \( C_p \):
- Line with the lowest slope (shallowest) corresponds to the monoatomic gas (M).
- Line with a medium slope corresponds to the diatomic gas (D).
- Line with the steepest slope corresponds to the polyatomic gas (P).

Thus, lines  a, b, and c correspond to  P, D, and M, respectively.

For an ideal gas, the relationship between \( C_p \) and \( C_v \) is:

\[
C_p - C_v = R
\]

where \( R \approx 2 \, \text{cal/mole-K} \).

Check each option:

1. If \( C_v = 3 \) and \( C_p = 5 \):
\[
C_p - C_v = 5 - 3 = 2 = R
\]
This matches the expected result.

2. Other sets will not satisfy \( C_p - C_v = 2 \) as accurately.

Thus, \( C_v = 3 \) and \( C_p = 5 \) is the most reliable set.

For a gas with \( \gamma = \frac{C_p}{C_v} = 1.4 \):

1. Atomicity: For diatomic gases, \( \gamma = 1.4 \) (common for diatomic molecules like \(\text{O}_2\), \(\text{N}_2\), etc.).

2. Heat capacities:
- \( C_v = \frac{R}{\gamma - 1} = \frac{R}{1.4 - 1} = \frac{5}{2} R \).
- \( C_p = \gamma C_v = 1.4 \times \frac{5}{2} R = \frac{7}{2} R \).

Thus, the atomicity is diatomic, and the values of \( C_p \) and \( C_v \) are \( \frac{7}{2} R \) and \( \frac{5}{2} R \), respectively.

According to the kinetic theory of gases:

- (A) Collisions are always elastic: Gas molecule collisions do not lose kinetic energy, so they are elastic.
- (B) There is no force of attraction among the molecules: Assumption of ideal gases is no intermolecular forces.
- (C) Only a small number of molecules have very high velocity: Most molecules have moderate speeds; only a few have very high speeds.
- (D) Between collisions, the molecules move in straight lines with constant velocities: Molecules move with constant speed in straight lines until they collide.

All options are correct as per the assumptions of kinetic theory.

The root mean square (rms) velocity of a gas is given by:

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

where \( M \) is the molar mass. Since all gases are at the same temperature, the gas with the smallest molar mass will have the highest \( v_{\text{rms}} \).

Hydrogen has the smallest molar mass among the options, so it has the maximum rms velocity.

For 15 g of nitrogen gas (N₂), where the molar mass \( M = 28 \, \text{g/mol} \):

1. Calculate the number of moles, \( n = \frac{\text{mass}}{\text{molar mass}} = \frac{15}{28} \).

2. Using the ideal gas law \( PV = nRT \):

\[
PV = \frac{15}{28} RT
\]

So, the equation of state is \( PV = \frac{15}{28} RT \).

According to Newton's law of cooling, the rate of cooling is proportional to the temperature difference between the object and its surroundings. As the liquid cools, the temperature difference between the liquid and the surroundings decreases, which slows down the rate of cooling.

Therefore, it will take greater than 5 minutes to cool from 60°C to 50°C, as the temperature difference is smaller, resulting in a slower cooling rate.