Physics MCQ, Ideal Gas Law, pV = constant, Thermodynamics, Physics Questions, Work Done by Gas, Detailed Solutions

Q1: One mole of an ideal gas at a temperature \( T_1 \) expands slowly according to the law \( pV = \text{constant} \). Its final temperature is \( T_2 \). The work done by the gas is:

1. \( R(T_2 - T_1) \)
2. \( 2R(T_2 - T_1) \)
3. \( \frac{R}{2}(T_2 - T_1) \)
4. \( \frac{3R}{2}(T_2 - T_1) \)

Solution:

For an isothermal process, the work done is given by:

\[ W = \int_{V_1}^{V_2} p \, dV \]

From the ideal gas law, \( pV = nRT \), where \( p = \frac{nRT}{V} \). Substituting:

\[ W = nR \int_{V_1}^{V_2} \frac{T \, dV}{V} \]

For the given process, since \( T \propto V \), \( T = T_1 \frac{V}{V_1} \). Integrating:

\[ W = R(T_2 - T_1) \]

Correct Answer: (A) \( R(T_2 - T_1) \)

Q2: A particle moves along the parabolic path \( y = ax^2 \) in such a way that the \( y \)-component of the velocity remains constant, say \( c \). The \( x \)- and \( y \)-coordinates are in meters. Then the acceleration of the particle at \( x = 1 \, \text{m} \) is:

1. \( a c \, \hat{k} \)
2. \( 2 a c^2 \, \hat{j} \)
3. \( - \frac{a^4 c^2}{2} \, \hat{i} \)
4. \( - \frac{a^2 c}{2} \, \hat{i} \)

Solution:

The path of the particle is \( y = ax^2 \). Differentiating this with respect to \( t \):

\[ \frac{dy}{dt} = c \quad \text{(since the \( y \)-component of velocity remains constant)} \]

Using the chain rule, \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \):

\[ c = 2ax \cdot \frac{dx}{dt} \]

\[ \therefore \frac{dx}{dt} = \frac{c}{2ax} \]

Differentiating \( \frac{dx}{dt} \) again with respect to \( t \) gives the acceleration in the \( x \)-direction:

\[ a_x = \frac{d}{dt} \left( \frac{c}{2ax} \right) \]

Using the chain rule, this becomes:

\[ a_x = -\frac{c}{2ax^2} \cdot \frac{dx}{dt} = -\frac{c}{2ax^2} \cdot \frac{c}{2ax} = -\frac{a^2c}{2} \]

The acceleration vector is along \( \hat{i} \) with a magnitude:

\[ a_x = -\frac{a^2c}{2} \, \hat{i} \]

Correct Answer: (D) \( -\frac{a^2c}{2} \, \hat{i} \)

Q1: One mole of an ideal gas at a temperature \( T_1 \) expands slowly according to the law \( pV = \text{constant} \). Its final temperature is \( T_2 \). The work done by the gas is:

Q2: A particle moves along the parabolic path \( y = ax^2 \) in such a way that the \( y \)-component of the velocity remains constant, say \( c \). The \( x \)- and \( y \)-coordinates are in meters. Then the acceleration of the particle at \( x = 1 \, \text{m} \) is: