Question 71:
easyAssertion (A): Molar heat capacity of an ideal monoatomic gas at constant volume is a constant at all temperatures.
Reason (R): As the temperature of an monoatomic ideal gas is increased, number of degrees of freedom of gas molecules remains constant.
For an ideal monoatomic gas, \(C_v = \frac{3}{2}R\) as it only has 3 translational degrees of freedom. This number \(f=3\) remains constant with temperature. Thus, \(C_v\) is constant. Reason (R) correctly explains Assertion (A).
Question 72:
easyAssertion (A): Experimental results indicate that the molar specific heat of hydrogen gas at constant volume below \( 50 \text{ K} \) is equal to \( 5/2 R \), where \( R \) is the universal gas constant.
Reason (R): A diatomic hydrogen molecule possesses three translational and two rotational degrees of freedom at all temperatures.
Assertion (A) is false. Below \( 50 \text{ K} \), hydrogen's rotational modes freeze out, so \( C_V \) approaches \( 3/2 R \), not \( 5/2 R \).
Reason (R) is false because degrees of freedom depend on temperature; vibrational modes activate at high T, and rotational modes freeze out at low T.
Question 73:
easyAssertion (A): Molar heat capacity at constant pressure can be less than molar heat capacity at constant volume.
Reason (R): \( C_p – C_V = R \) is valid only for ideal monoatomic gas.
Assertion (A) is false; \( C_p \) is always greater than \( C_V \) because work is done at constant pressure.
Reason (R) is false; Mayer's relation, \( C_p - C_V = R \), is valid for all ideal gases, regardless of atomicity.
Question 74:
easyAssertion (A): An ideal gas is enclosed within a container fitted with a piston when volume of this enclosed gas is increased at constant temperature. The pressure exerted by the gas on the piston decreases.
Reason (R): In the above situation the rate of molecules striking the piston decreases. Therefore pressure exerted by gas on piston decreases.
Assertion (A) is true by Boyle's Law (\( PV = \text{constant} \) at constant \( T \)). Reason (R) explains (A) microscopically: increasing volume at constant temperature reduces the density of molecules and thus the frequency of collisions with the piston, leading to decreased pressure.
Question 75:
easyAssertion (A): A real gas behaves as an ideal gas at high temperature and low pressure.
Reason (R): At low pressure and high temperature intermolecular forces vanish away and volume of gas molecules is negligible.
Assertion (A) is true. Real gases approximate ideal gas behavior under conditions of high temperature (high kinetic energy overcomes intermolecular forces) and low pressure (molecules are far apart, making their own volume negligible).
Reason (R) accurately states these conditions as the underlying cause for ideal gas behavior. Thus, R is the correct explanation for A.
Question 76:
easyAssertion (A): On a V-T graph, the slope of an isobar increases with pressure.
Reason (R): At constant temperature, for an ideal gas its volume is directly proportional to its pressure.
For an isobar, \( V = (\frac{nR}{P})T \). The slope on a V-T graph is \( \frac{nR}{P} \). As P increases, slope decreases, so (A) is false. Boyle's law states that at constant T, \( V \propto \frac{1}{P} \), i.e., V is inversely proportional to P, so (R) is false.
Question 77:
easyAssertion (A): For an ideal gas, at constant temperature, the product of the pressure and volume is constant.
Reason (R): The mean square velocity of gas molecules is inversely proportional to mass of molecule.
Boyle's Law states that for an ideal gas at constant T, \( PV = \text{constant} \). So (A) is true. The mean square velocity \( = \frac{3kT}{m} \), so it is inversely proportional to molecular mass m.
So (R) is true. However, (R) does not explain Boyle's law (A).
Question 78:
easyIn ideal condition, the maximum efficiency that can be derived from a heat engine built operating between \(600\text{ K}\) reservoir and \(200\text{ K}\) sink, is
The maximum efficiency is given by the Carnot efficiency formula: \(\eta = 1 - \frac{T_2}{T_1}\). Here, \(T_1 = 600\text{ K}\) and \(T_2 = 200\text{ K}\), so \(\eta = 1 -\frac{200}{600} = 1 - \frac{1}{3} = \frac{2}{3} \approx 66.67%\).
Question 79:
easyFor \(n\) mole of an ideal gas, the correct equation of \(1^{\text{st}}\) law of thermodynamics corresponding to isobaric process will be (symbols have their usual meanings)
According to the first law of thermodynamics, \(Q = \Delta U + W\). For an isobaric process, the work done is \(W = P\Delta V = nR\Delta T\). Therefore, both equations (1) and (2) are correct representation.
Question 80:
easyConsider the following thermodynamic parameters:
(a) Heat
(b) Internal energy
(c) Work
Which of the given parameters are path functions?
Heat and work depend on the path taken by the system during a thermodynamic process, making them path functions. Internal energy is a state function as it depends only on the initial and final states of the system.