Solution:

The internal energy \( U \) of an ideal gas is given by:

\[
U = n \cdot C_V \cdot T
\]

For a diatomic gas like hydrogen (\( \text{H}_2 \)), \( C_V = \frac{5}{2} R \), and for a monoatomic gas like helium (\( \text{He} \)), \( C_V = \frac{3}{2} R \).

Given that the internal energy of \( n_1 \) moles of hydrogen at temperature \( T \) is equal to the internal energy of \( n_2 \) moles of helium at temperature \( 2T \), we have:

\[
n_1 \cdot \frac{5}{2} R \cdot T = n_2 \cdot \frac{3}{2} R \cdot (2T)
\]

Simplifying:

\[
\frac{5}{2} n_1 = 3 n_2
\]

Rearrange to find the ratio \( \frac{n_1}{n_2} \):

\[
\frac{n_1}{n_2} = \frac{3}{5} \cdot \frac{2}{1} = \frac{6}{5}
\]

Answer: The ratio \( \frac{n_1}{n_2} \) is \( \frac{6}{5} \).

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Solution:

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 \( 9 P_1 V_1 \).

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Solution:

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

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

where \( m \) is the molar mass of the gas. Since all gases are at the same temperature, \( v_{\text{rms}} \) is inversely proportional to the square root of their molar mass:

\[
v_{\text{rms}} \propto \frac{1}{\sqrt{m}}
\]

Among air, oxygen (\( \text{O}_2 \)), carbon dioxide (\( \text{CO}_2 \)), and hydrogen (\( \text{H}_2 \)), hydrogen has the lowest molar mass. Therefore, it has the highest r.m.s. velocity.

Answer: Hydrogen has the maximum r.m.s. velocity.

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Solution:

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

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

For nitrogen (\( \text{N}_2 \)) and oxygen (\( \text{O}_2 \)) gases to have the same \( v_{\text{rms}} \):

\[
\sqrt{\frac{3 k_B T_{\text{N}_2}}{M_{\text{N}_2}}} = \sqrt{\frac{3 k_B T_{\text{O}_2}}{M_{\text{O}_2}}}
\]

Squaring both sides and simplifying:

\[
\frac{T_{\text{N}_2}}{T_{\text{O}_2}} = \frac{M_{\text{N}_2}}{M_{\text{O}_2}}
\]

Given:
- \( T_{\text{O}_2} = 127^\circ \text{C} = 127 + 273 = 400 \, \text{K} \),
- Molar masses \( M_{\text{N}_2} = 28 \) and \( M_{\text{O}_2} = 32 \).

Substitute values:

\[
T_{\text{N}_2} = \frac{28}{32} \times 400 = 350 \, \text{K}
\]

Convert \( 350 \, \text{K} \) to Celsius:

\[
350 - 273 = 77^\circ \text{C}
\]

Answer: The temperature is \( 77^\circ \text{C} \).

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Solution:

At Normal Temperature and Pressure (NTP), 1 mole of any ideal gas occupies 22.4 liters and contains Avogadro's number (\(6.022 \times 10^{23}\)) of molecules.

Since each gas sample has the same volume (1 cm³) and is at NTP, they all contain the same number of molecules. This is because, at the same conditions of temperature and pressure, equal volumes of gases have equal numbers of molecules (Avogadro's Law).

Answer: All samples have the same number of molecules.

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Solution:

The relation between the root mean square (r.m.s.) velocity \( v_{\text{rms}} \) and the most probable velocity \( v_{\text{mp}} \) of a gas is derived from their respective formulas:

1. **Most probable velocity** \( v_{\text{mp}} \):
\[
v_{\text{mp}} = \sqrt{\frac{2 k_B T}{m}}
\]

2. **Root mean square velocity** \( v_{\text{rms}} \):
\[
v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}
\]

Dividing \( v_{\text{rms}} \) by \( v_{\text{mp}} \):

\[
\frac{v_{\text{rms}}}{v_{\text{mp}}} = \sqrt{\frac{3}{2}}
\]

Thus:

\[
v_{\text{rms}} = \sqrt{\frac{3}{2}} \, v_{\text{mp}}
\]

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Solution:

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 \( 4.24 \).

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Solution:

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

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

Since \( v_{\text{rms}} \propto \sqrt{T} \), if the temperature changes, the r.m.s. speed changes proportionally to the square root of the temperature.

Let the initial r.m.s. speed at \( T = 120 \, \text{K} \) be \( v \). Then, at \( T = 480 \, \text{K} \), the r.m.s. speed \( v' \) is:

\[
v' = v \cdot \sqrt{\frac{480}{120}} = v \cdot \sqrt{4} = 2v
\]

So, the r.m.s. speed at \( 480 \, \text{K} \) will be \( 2v \).

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Solution:

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 \( \text{H}_2 \).

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Solution:

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:

\[
0.5n_1 = 0.5n_2
\]

\[
n_1 = n_2
\]

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