The total average kinetic energy \( E \) of translatory motion for an ideal gas is given by:

\[
E = \frac{3}{2} nRT
\]

where:
- \( n \) is the number of moles,
- \( R \) is the gas constant (\( 8.314 \, \text{J/mol·K} \)),
- \( T \) is the temperature in Kelvin.

We can rearrange to find \( nRT \):

\[
nRT = \frac{2}{3} E
\]

The ideal gas law also gives us:

\[
PV = nRT
\]

Thus,

\[
P = \frac{nRT}{V} = \frac{\frac{2}{3} E}{V}
\]

Substitute \( E = 1.5 \times 10^5 \, \text{J} \) and \( V = 20 \, \text{litres} = 20 \times 10^{-3} \, \text{m}^3 \):

\[
P = \frac{\frac{2}{3} \times 1.5 \times 10^5}{20 \times 10^{-3}}
\]

\[
P = \frac{10^5}{20 \times 10^{-3}} = 5 \times 10^6 \, \text{N/m}^2
\]

So, the pressure is \( 5 \times 10^6 \, \text{N/m}^2 \).

Thank you for the clarification. Given the correct answer, let's interpret the graph accordingly.

In this pressure (\( P \)) vs. temperature (\( T \)) graph:

1. **Parallel Lines**: When two lines have the same slope, it implies they have the same volume. This is because the slope in a \( P \)-\( T \) graph (for constant \( V \)) is given by \( \frac{nR}{V} \).

- Since lines **1** and **2** are parallel, we have \( V_1 = V_2 \).
- Similarly, lines **3** and **4** are parallel, so \( V_3 = V_4 \).

2. **Comparing Slopes**: The line with a steeper slope corresponds to a smaller volume, and the line with a shallower slope corresponds to a larger volume.

- Since lines 1 and 2 have a shallower slope compared to lines 3 and 4, we conclude that \( V_1 = V_2 > V_3 = V_4 \).

Final Answer:
- \( V_1 = V_2 \), \( V_3 = V_4 \), and \( V_2 > V_3 \).

For a monoatomic gas under isobaric (constant pressure) conditions, the heat required \( Q \) is given by:

\[
Q = n C_p \Delta T
\]

where:
- \( n = 2 \) moles,
- \( C_p = \frac{5}{2} R \) (specific heat at constant pressure for monoatomic gas),
- \( \Delta T = 50^\circ \text{C} - 0^\circ \text{C} = 50 \) K.

Substitute values:

\[
Q = 2 \times \frac{5}{2} R \times 50 = 250R
\]

So, the answer is \( 250R \).

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:

\[
0.5n_1 = 0.5n_2
\]

\[
n_1 = n_2
\]

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.

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 \(200 \, \text{K}\).

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.