Assertion (A): It is possible for both the pressure and volume of a monoatomic ideal gas of a given amount to change simultaneously without causing the internal energy of the gas to change.
Reason (R): The internal energy of an ideal gas of a given amount remains constant if temperature does not change. It is possible to have a process in which pressure and volume are changed such that temperature remains constant.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
For an ideal gas, internal energy \(U\) depends only on temperature \(T\). If \(U\) is constant, then \(T\) is constant. For a constant temperature process (isothermal), pressure \(P\) and volume \(V\) can change while \(T\) remains constant (as \(PV = nRT\)). Thus, both assertion and reason are true, and the reason correctly explains the assertion.
Assertion (A): Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion adiabatically.
Reason (R): Temperature remains constant in isothermal expansion but not in adiabatic expansion.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A) is true. For the same volume expansion, work done \(W = int P dV\). In isothermal expansion, \(P\) drops slower than in adiabatic expansion (due to heat supply), so the area under the \(P-V\) curve is greater for isothermal. Reason (R) is also true and explains why \(P\) behaves differently, leading to different work done.
Assertion (A): During the melting of a slab of ice at \(273\text{ K}\) at \(1\text{ atm}\) positive work is done on the ice-water system by the atmosphere.
Reason (R): In above process, the internal energy of ice-water system increases.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
When ice melts to water, its volume decreases \(\Delta V < 0\). Work done *by* the atmosphere *on* the system is \(-P\Delta V\), which is positive. So (A) is true. During melting, latent heat is absorbed, increasing internal energy \(\Delta U = Q - W\). Since \(Q\) is positive and \(W\) (work by system) is negative, \(\Delta U\) is positive. So (R) is true. However, the increase in internal energy is not the reason for the work done by the atmosphere; it's the volume change. So (R) does not explain (A).
Assertion (A): It is possible for both the pressure and volume of a monoatomic ideal gas of a given amount to change simultaneously without causing the internal energy of the gas to change.
Reason (R): The internal energy of an ideal gas of a given amount remains constant if temperature does not change. It is possible to have a process in which pressure and volume are changed such that temperature remains constant.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Concept: Internal energy of an ideal gas depends only on temperature.
Formula: For ideal gas, \( U = f(T) \). For monoatomic, \( U = \frac{3}{2} nRT \). Isothermal process implies \( T \) is constant.
Solution: If \( U \) is constant, then \( T \) is constant. An isothermal process allows simultaneous change in \( P \) and \( V \) while \( T \) (and thus \( U \)) remains constant. Reason correctly explains this.
Assertion (A): Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion adiabatically.
Reason (R): Temperature remains constant in isothermal expansion but not in adiabatic expansion.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Concept: Work done is area under P-V curve. Isothermal vs. Adiabatic expansion.
Formula: \( W = \int P dV \). Isothermal: \( PV = \text{constant} \). Adiabatic: \( PV^{\gamma} = \text{constant} \), where \( \gamma > 1 \).
Solution: During expansion from the same initial state to the same final volume, the pressure in isothermal process drops slower than in adiabatic process, leading to more work done. Temperature remains constant in isothermal and changes in adiabatic.
Assertion (A): Bursting of balloon is not a equilibrium state.
Reason (R): Equilibrium state of a thermodynamic system is completely described by specific values of some macroscopic properties.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Concept: Thermodynamic equilibrium. A bursting balloon is a spontaneous, non-equilibrium process. An equilibrium state is characterized by constant macroscopic properties. Both assertion (A) and reason (R) are true, and (R) correctly explains (A).
Assertion (A): Work and heat both can be converted into each other in any condition.
Reason (R): Work and Heat both are different form of energy.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Concept: First Law of Thermodynamics and nature of work/heat. Work and heat are forms of energy transfer, not different forms of energy itself. Their interconversion is governed by thermodynamic laws and not possible under 'any condition'. Both (A) and (R) are false.
Assertion (A): If volume of a gas is increasing but temperature of the gas is decreasing, then heat given to the gas may be positive, negative or zero.
Reason (R): Heat given to a gas is a path function, it is not a state function.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Concept: First Law of Thermodynamics (( Delta U = Q - W )). If volume increases, (W > 0) (work done by gas). If temperature decreases, ( Delta U < 0 ). So, ( Q = Delta U + W ) can be positive, negative, or zero. Heat is indeed a path function. Both (A) and (R) are true, and (R) explains (A).
Assertion (A): The area of entropy versus temperature graph of a cyclic process, is equal to work done.
Reason (R): Change in internal energy of cyclic process is zero.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Concept: T-S diagram and cyclic processes. The area enclosed by a (T-S) diagram for a cyclic process represents the net heat exchanged, ( Q_{net} ), not the work done. For a cyclic process, ( Delta U = 0 ) is true, but Assertion (A) is false. Therefore, both (A) and (R) are false in relation to the explanation.
Assertion (A): On sudden expansion a gas cools.
Reason (R): On sudden expansion, no heat is supplied to system and hence gas does work at the expense of its internal energy.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Concept: Adiabatic expansion and First Law of Thermodynamics. Sudden expansion is a rapid process, approximated as adiabatic (( Q = 0 )). The gas does work ( W > 0 ). By \( \Delta U = Q - W ), ( \Delta U \) becomes negative, leading to a decrease in internal energy and thus cooling. Both (A) and (R) are true, and (R) explains (A).