Assertion (A): In adiabatic compression, the temperature of system gets decreased.
Reason (R): Adiabatic compression is a slow process.
In adiabatic compression, work is done on the gas \(W =0\), leading to an increase in temperature. Thus (A) is false. Adiabatic processes are typically rapid to prevent heat exchange. Thus (R) is false. Both (A) and (R) are false.
Assertion (A): All processes in which P and V are proportional, take place at constant temperature.
Reason (R): Work done in a thermodynamical process is path independent.
If \(P propto V\) (i.e., \(P=kV\)), then for an ideal gas \(PV=nRT\) implies \(kV^2=nRT\), so \(T \propto V^2\). Thus, temperature is not constant. (A) is false. Work done \(W = \int PdV\) depends on the path taken on a \(P-V\) diagram, so it is path dependent. (R) is false. Both (A) and (R) are false.
Assertion (A): During adiabatic expansion of an ideal gas, temperature falls but entropy remains constant.
Reason (R): During adiabatic expansion, work is done by the gas using a part of internal energy and no heat exchange takes place the system and the surrounding.
During adiabatic expansion, \(Q=0\). The gas does work by using its internal energy, causing temperature to fall. For a reversible adiabatic process, entropy \(\Delta S=0\). (A) is true. (R) correctly explains the energy changes (no heat exchange, internal energy conversion to work) that lead to temperature drop and, for reversible processes, constant entropy. Both are true and (R) is the correct explanation of (A).
Assertion (A): In a free adiabatic expansion of an ideal gas, the final state is the same as the initial state.
Reason (R): As temperature of a gas increases work done by it is positive.
For free adiabatic expansion, \( Q = 0 \) and \( W = 0 \). According to the first law of thermodynamics, \( \Delta U = Q - W = 0 \). For an ideal gas, \( \Delta U = nC_v\Delta T \), so \( \Delta T = 0 \), meaning the final temperature is the same as the initial temperature. However, the final state (P,V,T) is not entirely the same as the initial state, only T is.
Thus, Assertion (A) is false. If temperature increases, work done by the gas can be positive, but it is not a direct consequence, and expansion (positive work) typically cools the gas. So, Reason (R) is also false. Therefore, both (A) and (R) are false.
Assertion (A): In adiabatic process, work done on the system is equal to negative of change in internal energy.
Reason (R): In adiabatic process change of heat zero.
In an adiabatic process, the system is perfectly insulated so that no heat exchange occurs \(Q = 0\). According to the First Law of Thermodynamics \(\Delta U = Q - W_{by}\), this means any change in internal energy is due entirely to work: \(\Delta U = -W_{by}\)
Assertion (A): In cyclic process change in internal energy is zero.
Reason (R): In cyclic process net work done is zero.
For a cyclic process, the system returns to its initial state. Internal energy \( U \) is a state function, meaning its change depends only on the initial and final states. Since the initial and final states are the same in a cyclic process, the change in internal energy \( \Delta U \) is zero. Thus, Assertion (A) is true. According to the first law of thermodynamics, \( \Delta U = Q - W \). Since \( \Delta U = 0 \), it implies \( Q = W \). For a cyclic process, net work done \( W \) is generally not zero (e.g., heat engines operate in cycles to produce net work). Thus, Reason (R) is false. Therefore, (A) is true but (R) is false.
Assertion (A): State variables (P, V and T) of any gas at low densities obey the equation \( PV = nRT \).
Reason (R): Real gases are good approximation of an ideal gas at low density.
The ideal gas equation \( PV = nRT \) describes the behavior of an ideal gas. Real gases approximate ideal gas behavior at low densities (and high temperatures) because intermolecular forces become negligible and the volume of gas molecules is tiny compared to the total volume.
Thus, Assertion (A) is true, and Reason (R) is true. Reason (R) directly explains why real gases obey the ideal gas equation at low densities. Therefore, (R) is the correct explanation of (A).
Assertion (A): An ideal gas expands isothermally, during this process, it absorbs \( 25 \text{ J} \) heat. In the first law of thermodynamics, work done on the gas will be \( -25 \text{ J} \).
Reason (R): There will be no change in the internal energy of the gas during isothermal expansions.
For an ideal gas undergoing an isothermal process, the temperature remains constant (\( Delta T = 0 \)). Since the internal energy of an ideal gas depends only on temperature, the change in internal energy \( Delta U = 0 \). Thus, Reason (R) is true. From the first law of thermodynamics, \( Delta U = Q - W_{by} \), where \( W_{by} \) is work done by the gas. Since \( Delta U = 0 \), we have \( Q = W_{by} \). Given that the gas absorbs \( 25 text{ J} \) heat, \( Q = +25 text{ J} \). Therefore, work done by the gas is \( W_{by} = +25 text{ J} \). Work done on the gas is \( W_{on} = -W_{by} = -25 text{ J} \). This matches Assertion (A), so (A) is true. Reason (R) correctly explains (A) because the condition \( Delta U = 0 \) is central to deriving the relationship between heat and work in this process.
Assertion (A): There is no change in internal energy for ideal gas at constant temperature.
Reason (R): Internal energy of an ideal gas is a function of temperature only.
For an ideal gas, internal energy \(U\) depends solely on temperature \(T\). Therefore, if \(T\) is constant, \(Delta U = 0\). Reason (R) correctly explains Assertion (A).
Assertion (A): At \(0K\), pressure of an ideal gas becomes zero.
Reason (R): At \(0K\), according to ideal gas equation \(PV = 0\), volume cannot be zero hence pressure should be zero to satisfy this equation.
From the ideal gas equation \(PV = nRT\), if \(T = 0K\), then \(PV = 0\). Since volume \(V\) cannot be zero, pressure \(P\) must be zero. Reason (R) directly explains Assertion (A).