NEET Thermal Physics MCQs & PYQs

In the gas equation \( PV = nRT \), the universal gas constant \( R \) has a fixed value but its numerical value depends on the units of \( P \), \( V \), and \( T \).

Answer: The value of the universal gas constant depends only on the units of measurement.

The expression \(\frac{PV}{kT}\) can be analyzed using the ideal gas law:

\[
PV = NkT
\]

where:
- \( P \) is pressure,
- \( V \) is volume,
- \( N \) is the number of molecules,
- \( k \) is Boltzmann's constant, and
- \( T \) is temperature.

Rearranging, we get:

\[
\frac{PV}{kT} = N
\]

Answer: This quantity represents the "number of molecules in the gas".

Given:

- Initial temperature, \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \)
- Final temperature, \( T_2 = 87^\circ \text{C} = 360 \, \text{K} \)
- Initial pressure, \( P_1 = 2 \, \text{atm} \)

Using the pressure-temperature relationship for a gas at constant volume:

\[
\frac{P_2}{P_1} = \frac{T_2}{T_1}
\]

Solving for \( P_2 \):

\[
P_2 = P_1 \times \frac{T_2}{T_1} = 2 \times \frac{360}{300} = 2 \times 1.2 = 2.4 \, \text{atm}
\]

Answer: \( P_2 = 2.4 \, \text{atm} \)

Given:

- Initial temperature, \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \)
- We want to expel half the mass of air, meaning the final mass, \( m_2 = \frac{m_1}{2} \).

Using the ideal gas law, \( PV = nRT \), and since pressure and volume are constant, \( \frac{m}{T} = \text{constant} \).

Thus,

\[
\frac{m_1}{T_1} = \frac{m_2}{T_2}
\]

Substitute \( m_2 = \frac{m_1}{2} \):

\[
\frac{m_1}{300} = \frac{\frac{m_1}{2}}{T_2}
\]

Solving for \( T_2 \):

\[
T_2 = 2 \times 300 = 600 \, \text{K}
\]

Converting back to Celsius:

\[
T_2 = 600 - 273 = 327^\circ \text{C}
\]

Answer: \( T_2 = 327^\circ \text{C} \)

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

To analyze the \( V \) versus \( T \) curves for an ideal gas at constant pressures \( P_1 \) and \( P_2 \):

According to Charles's Law:

\[
\frac{V}{T} = \frac{nR}{P}
\]

This implies:

1. At constant pressure, the volume \( V \) is directly proportional to the temperature \( T \), resulting in a straight line.

2. For a given temperature, the volume \( V \) will be greater at a lower pressure \( P \), because \( V \propto \frac{1}{P} \) for a fixed amount of gas.

In the graph:
- The line with a steeper slope corresponds to a lower pressure (since a lower \( P \) results in a larger \( V \) for the same \( T \)).

Since \( P_2 \) has a steeper slope than \( P_1 \), we conclude that:

\[
P_1 > P_2
\]

Therefore, the correct answer is \( P_1 > P_2 \).

When a metal rod is heated, its atoms gain kinetic energy and vibrate more vigorously. As they vibrate, they tend to move slightly further apart because the increased energy weakens the attractive forces that hold them at a fixed distance. This increased atomic spacing results in the rod expanding in size.

Thus, the expansion of the metal rod occurs because the distance among its atoms increases with temperature. This phenomenon is the essence of thermal expansion.