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 \( T' = 2T \).

Let the temperature of the surroundings be \( T_s \).

Given:
- First cooling: from \( 62^\circ \text{C} \) to \( 50^\circ \text{C} \) in 10 minutes.
- Second cooling: from \( 50^\circ \text{C} \) to \( 42^\circ \text{C} \) in the next 10 minutes.

Step 1: Use Newton's Law of Cooling
According to Newton's law:
\[
\frac{T_1 - T_2}{T_1 - T_s} = \frac{T_2 - T_3}{T_2 - T_s}
\]

Step 2: Substitute Values
Let \( T_1 = 62^\circ \text{C} \), \( T_2 = 50^\circ \text{C} \), and \( T_3 = 42^\circ \text{C} \).

Then:
\[
\frac{62 - 50}{62 - T_s} = \frac{50 - 42}{50 - T_s}
\]

Step 3: Solve for \( T_s \)
\[
\frac{12}{62 - T_s} = \frac{8}{50 - T_s}
\]

Cross-multiplying:
\[
12(50 - T_s) = 8(62 - T_s)
\]

Expanding and simplifying:
\[
600 - 12T_s = 496 - 8T_s
\]

\[
4T_s = 104 \Rightarrow T_s = 26^\circ \text{C}
\]

So, the temperature of the surroundings is \( 26^\circ \text{C} \).

Using Newton's law of cooling:

Given:
- Initial cooling: from \( T_1 = 61^\circ \text{C} \) to \( T_2 = 59^\circ \text{C} \) in \( \Delta t_1 = 10 \) minutes, with room temperature \( T_s = 30^\circ \text{C} \).
- New cooling required: from \( T_3 = 51^\circ \text{C} \) to \( T_4 = 49^\circ \text{C} \).

Using the proportionality from Newton's law:
\[
\frac{\Delta t_1}{\Delta t_2} = \frac{\text{Average Temp Difference in Initial Cooling}}{\text{Average Temp Difference in New Cooling}}
\]

Calculate average temperature differences:
1. For initial cooling: \( \frac{61 + 59}{2} - 30 = 60 - 30 = 30 \)
2. For new cooling: \( \frac{51 + 49}{2} - 30 = 50 - 30 = 20 \)

Then:
\[
\frac{10}{\Delta t_2} = \frac{30}{20}
\]

Solving for \( \Delta t_2 \):
\[
\Delta t_2 = 10 \times \frac{20}{30} = 15 \text{ minutes}
\]

Using Newton's law of cooling:

Given:
- Initial cooling: \( T_1 = 80^\circ \text{C} \) to \( T_2 = 79.9^\circ \text{C} \) in 5 min, surroundings \( T_s = 20^\circ \text{C} \).
- New cooling required: from \( 70^\circ \text{C} \) to \( 69.9^\circ \text{C} \).

For both cases, Newton's law states:
\[
\frac{\Delta T}{\Delta t} \propto (T - T_s)
\]

Since the temperature difference is small in both cases, we assume:
\[
\frac{\Delta t_1}{\Delta t_2} = \frac{T_1 - T_s}{T_3 - T_s}
\]

So:
\[
\frac{5}{\Delta t_2} = \frac{80 - 20}{70 - 20} \Rightarrow \Delta t_2 = 5 \times \frac{70 - 20}{80 - 20} = 5 \times \frac{50}{60} = 6 \text{ min}
\]

Using Newton's law of cooling:

Let \( T_s \) be the temperature of the surroundings, and \( T_1 = 70^\circ \text{C} \), \( T_2 = 60^\circ \text{C} \), \( T_3 = 54^\circ \text{C} \).

The average temperatures in each time interval are:
1. For first 5 minutes: \( \frac{70 + 60}{2} = 65^\circ \text{C} \)
2. For next 5 minutes: \( \frac{60 + 54}{2} = 57^\circ \text{C} \)

According to Newton's law:
\[
\frac{T_1 - T_2}{T_1 - T_s} = \frac{T_2 - T_3}{T_2 - T_s}
\]

Substitute values:
\[
\frac{70 - 60}{65 - T_s} = \frac{60 - 54}{57 - T_s}
\]

Simplifying:
\[
\frac{10}{65 - T_s} = \frac{6}{57 - T_s}
\]

Cross-multiplying and solving for \( T_s \) gives \( T_s = 45^\circ \text{C} \).

To solve this, we use Wien's Displacement Law, which states:

\[
\lambda_{\text{max}} = \frac{b}{T}
\]

Where:
- \(\lambda_{\text{max}}\) is the wavelength at which the maximum energy is emitted,
- \(b = 2.88 \times 10^6 \, \text{nmK}\) is Wien's constant,
- \(T = 5760 \, K\) is the temperature of the black body.

Step 1: Calculate \(\lambda_{\text{max}}\)
\[
\lambda_{\text{max}} = \frac{2.88 \times 10^6}{5760} = 500 \, \text{nm}
\]

This means that the maximum energy is emitted at a wavelength of 500 nm.

 Step 2: Compare the energies at different wavelengths

- At 500 nm (\( \lambda_{\text{max}} \)): This is where the black body emits the maximum energy. So, \(U_2\) will be the largest.
- At 250 nm: This wavelength is shorter than \(\lambda_{\text{max}}\), and the energy decreases as we move away from the peak, so \(U_1 < U_2\).
- At 1000 nm: This wavelength is longer than \(\lambda_{\text{max}}\), and the energy decreases further, so \(U_3 < U_2\).

Thus, \(U_2 > U_1\), and the correct comparison is \(U_2 > U_1\).