Using the ideal gas law \( PV = nRT \), we find the moles \( n \) in each vessel:

1. For vessel \( A \) with \( O_2 \):
\[
n_{\text{O}_2} = \frac{PV}{RT}
\]

2. For vessel \( B \) with \( H_2 \) at \( 2P \):
\[
n_{\text{H}_2} = \frac{2PV}{RT} = 2 \times \frac{PV}{RT}
\]

Now, the mass \( m = n \times \text{molar mass} \):

- Mass of \( O_2 \) in \( A = n_{\text{O}_2} \times 32 = \frac{PV}{RT} \times 32 \)
- Mass of \( H_2 \) in \( B = n_{\text{H}_2} \times 2 = 2 \times \frac{PV}{RT} \times 2 \)

So, the mass ratio is:
\[
\frac{\text{mass of } O_2}{\text{mass of } H_2} = \frac{\frac{PV}{RT} \times 32}{2 \times \frac{PV}{RT} \times 2} = \frac{32}{4} = 8:1
\]

For one mole of an ideal gas, we use the ideal gas law:

\[
PV = RT
\]

Given \( P = \frac{P_0}{1 + \left( \frac{V_0}{V} \right)^2} \), find \( T \) at \( V = V_0 \) and \( V = 2V_0 \).

1. When \( V = V_0 \):
\[
P = \frac{P_0}{1 + \left( \frac{V_0}{V_0} \right)^2} = \frac{P_0}{1 + 1} = \frac{P_0}{2}
\]
\[
T_1 = \frac{PV}{R} = \frac{\frac{P_0}{2} \cdot V_0}{R} = \frac{P_0 V_0}{2R}
\]

2. When \( V = 2V_0 \):
\[
P = \frac{P_0}{1 + \left( \frac{V_0}{2V_0} \right)^2} = \frac{P_0}{1 + \frac{1}{4}} = \frac{P_0}{\frac{5}{4}} = \frac{4P_0}{5}
\]
\[
T_2 = \frac{PV}{R} = \frac{\frac{4P_0}{5} \cdot 2V_0}{R} = \frac{8P_0 V_0}{5R}
\]

3. Change in temperature:
\[
\Delta T = T_2 - T_1 = \frac{8P_0 V_0}{5R} - \frac{P_0 V_0}{2R} = \frac{11P_0 V_0}{10R}
\]

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 \( 9 P_1 V_1 \).

To solve this, we can set up a linear relationship between the actual Celsius scale (0ºC to 100ºC) and the thermometer's faulty scale (20ºC to 150ºC).

1. Set up the linear equation:

The faulty thermometer's scale can be represented as:
\[
T_{\text{faulty}} = a \cdot T_{\text{actual}} + b
\]

Using the freezing point:
\[
20 = a \cdot 0 + b \Rightarrow b = 20
\]

Using the boiling point:
\[
150 = a \cdot 100 + 20
\]
\[
130 = 100a \Rightarrow a = 1.3
\]

So, the relation is:
\[
T_{\text{faulty}} = 1.3 \cdot T_{\text{actual}} + 20
\]

2. Find the faulty reading at 60ºC actual temperature:
\[
T_{\text{faulty}} = 1.3 \cdot 60 + 20 = 78 + 20 = 98
\]

Therefore, the thermometer will read 98ºC at an actual temperature of 60ºC.

Given:

\[
P = \frac{A}{1 + \left( \frac{B}{V} \right)^2}
\]

Using the ideal gas equation, \( PV = nRT \), for initial and final states, we can express the temperature change.

 Step 1: Initial State (at \( V = B \))
\[
P_1 = \frac{A}{1 + \left( \frac{B}{B} \right)^2} = \frac{A}{2}
\]
\[
T_1 = \frac{P_1 V}{R} = \frac{\left(\frac{A}{2}\right) B}{R} = \frac{AB}{2R}
\]

Step 2: Final State (at \( V = 2B \))
\[
P_2 = \frac{A}{1 + \left( \frac{B}{2B} \right)^2} = \frac{A}{1 + \frac{1}{4}} = \frac{A}{\frac{5}{4}} = \frac{4A}{5}
\]
\[
T_2 = \frac{P_2 V}{R} = \frac{\left(\frac{4A}{5}\right) (2B)}{R} = \frac{8AB}{5R}
\]

 Step 3: Temperature Change
\[
\Delta T = T_2 - T_1 = \frac{8AB}{5R} - \frac{AB}{2R} = \frac{16AB - 5AB}{10R} = \frac{11AB}{10R}
\]

Answer: \(\frac{11AB}{10R}\)

Water has a unique property: it reaches its maximum density at 4°C. This means that as water is heated or cooled from 4°C, it expands. Here’s why:

1. If the water is heated above 4°C: It begins to expand, decreasing in density. Since the beaker is already full, this expansion causes water to overflow.

2. If the water is cooled below 4°C: Water also expands when cooled below 4°C due to the formation of an open, hexagonal structure in its molecules (especially as it approaches freezing). This expansion also leads to overflow.

Therefore, the beaker will overflow if the temperature changes from 4°C in either direction: heating above 4°C or cooling below 4°C.

The coefficient of apparent expansion of mercury in a vessel is given by:

\[
\gamma_{\text{apparent}} = \gamma_{\text{mercury}} - \alpha_{\text{vessel}}
\]

For the glass vessel:
\[
153 \times 10^{-6} = \gamma_{\text{mercury}} - \alpha_{\text{glass}}
\]

For the steel vessel:
\[
144 \times 10^{-6} = \gamma_{\text{mercury}} - \alpha_{\text{steel}}
\]

Subtracting these two equations, we get:

\[
153 \times 10^{-6} - 144 \times 10^{-6} = \alpha_{\text{steel}} - \alpha_{\text{glass}}
\]

Substitute \(\alpha_{\text{steel}} = 12 \times 10^{-6}/^\circ \text{C}\):

\[
9 \times 10^{-6} = 12 \times 10^{-6} - \alpha_{\text{glass}}
\]

Solving for \(\alpha_{\text{glass}}\):
\[
\alpha_{\text{glass}} = 3 \times 10^{-6}/^\circ \text{C}
\]