To solve this, we can use the ideal gas law in terms of the number of moles \( n \):

\[
PV = nRT
\]

Let:
- Initial moles of hydrogen be \( n_1 \),
- Final moles of hydrogen be \( n_2 \).

Given data:
- Initial pressure \( P \),
- Final pressure \( P/2 \),
- Initial temperature \( T_1 = 500 \, \text{K} \),
- Final temperature \( T_2 = 300 \, \text{K} \),
- Mass of hydrogen initially = 6 g.

Since \( n = \frac{PV}{RT} \), we can write the initial and final moles as:

\[
n_1 = \frac{PV}{RT_1} \quad \text{and} \quad n_2 = \frac{(P/2)V}{R \cdot 300}
\]

Taking the ratio \( \frac{n_2}{n_1} \):

\[
\frac{n_2}{n_1} = \frac{(P/2) \cdot V / (R \cdot 300)}{P \cdot V / (R \cdot 500)} = \frac{1}{2} \times \frac{500}{300} = \frac{5}{12}
\]

Since initial moles \( n_1 = \frac{6}{2} = 3 \) moles (using molar mass of H₂ = 2 g/mol), then final moles \( n_2 = \frac{5}{12} \times 3 = 2.5 \) moles.

Thus, moles leaked out = \( 3 - 2.5 = 0.5 \) moles, corresponding to \( 0.5 \times 2 = 1 \) gram of hydrogen.

So, 1 g of hydrogen leaks out.

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.

Using the ideal gas equation \(P = \frac{nRT}{V}\), with total moles \(n = 0.2 + 0.3 = 0.5\text{ mol}\) and volume \(V = 200 \times 10^{-6}\text{ m}^3\). Calculating gives \(P = \frac{0.5 \times 8.3 \times 200}{2 \times 10^{-4}} = 4.15 \times 10^6\text{ Pa}\).

For isothermal process in A: \(P_A = 2P_0\). For adiabatic process in B: \(P_B = P_0 (2)^\gamma\). The ratio of final pressures is \(\frac{P_B}{P_A} = \frac{P_0 2^\gamma}{2 P_0} = 2^{\gamma - 1}\).

Using the First Law of Thermodynamics, \(Delta Q = Delta U + W\). Here, \(Delta Q = 10\text{ J}\) and \(Delta U = -5\text{ J}\) (decrease). Substituting these values, \(10 = -5 + W\), which gives \(W = 15\text{ J}\).

The molar specific heat of a mixture at constant volume is \(C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}\). For monatomic helium, \(C_{v1} = 1.5R\), and for rigid diatomic hydrogen, \(C_{v2} = 2.5R\). Substituting gives \(C_{v,\text{mix}} = \frac{2(1.5R) + 3(2.5R)}{5} = 2.1R\).