Degrees of freedom of a rigid diatomic molecule is
A rigid diatomic molecule has 3 translational and 2 rotational degrees of freedom, giving a total of 5 degrees of freedom.
The equation of state for 14 g nitrogen gas at a pressure P and temperature T, when occupying a volume V will be
Number of moles \(n = \frac{m}{M} = \frac{14\text{ g}}{28\text{ g/mol}} = 0.5\text{ mol}\). Substituting \(n = \frac{1}{2}\) in \(PV = nRT\) gives \(PV = \frac{1}{2}RT\).
The unit of emissive power is
Emissive power is defined as the thermal energy radiated per unit area per unit time. Its SI unit is \(\text{J m}^{-2}\text{s}^{-1}\) (or \(\text{W m}^{-2}\)).
| Column I | Column II |
| (A) Total translational kinetic energy | (P) $\frac{5}{2} K_B T$ |
| (B) Total rotational kinetic energy | (Q) $nRT$ |
| (C) Total kinetic energy per mole | (R) $\frac{3}{2} nRT$ |
| (D) Total kinetic energy per molecule | (S) $\frac{5}{2} RT$ |
Translational KE is \(\frac{3}{2}nRT\). Rotational KE of a rigid diatomic gas is \(nRT\). Total KE per mole is \(\frac{5}{2}RT\) and total KE per molecule is \(\frac{5}{2}K_B T\).
Which of the following is not the correct assumption of kinetic theory of gases?
Kinetic theory assumes that gas molecules undergo continuous random motion and collide with each other as well as with the walls of the container.
A faulty thermometer shows $40^\circ\text{C}$ at ice point and $80^\circ\text{C}$ at steam point. The temperature at which its reading would be correct is
Using the relation $\frac{T - \text{LFP}}{\text{UFP} - \text{LFP}} = \frac{C - 0}{100 - 0}$, we substitute $T = C$ for correct reading. This gives $\frac{C - 40}{80 - 40} = \frac{C}{100}$, which simplifies to $C = \frac{200}{3} ^\circ\text{C}$.
A blackbody and a real body of identical dimensions are heated to same temperature. If ratio of rates of radiation of the blackbody and the real body is $4 : 3$, then emissivity of the real body is equal to
The rate of radiation is given by $E = e\sigma A T^4$. For a blackbody, $e = 1$. Given $\frac{E_b}{E} = \frac{4}{3} ⇒ \frac{1}{e} = \frac{4}{3}$, hence emissivity $e = 0.75$.
Water equivalent of a metallic bar is $100\text{ g}$, the energy required to increase its temperature by $1^\circ\text{C}$ is
Energy required is $Q = w \cdot \Delta T$, where $w$ is the water equivalent. Thus, $Q = 100\text{ g} \times 1\text{ cal/g}^\circ\text{C} \times 1^\circ\text{C} = 100\text{ cal}$.
For $n$ 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)
By first law, $Q = \Delta U + W$. In an isobaric process, the work done is $W = P\Delta V = nR\Delta T$. Therefore, both expressions are correct.
Consider 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 between states, whereas internal energy is a state function. Therefore, (a) and (c) are path functions.