Question 301:
easyIn ideal condition, the maximum efficiency that can be derived from a heat engine operating between $600\text{ K}$ reservoir and $200\text{ K}$ sink, is
Efficiency is given by $\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$. Substituting the given values: $\eta = 1 -\frac{200}{600} =\frac{2}{3} \approx 66.67%$.
Question 302:
easyThe equation of state for \(14\text{ g}\) nitrogen gas at a pressure \(P\) and temperature \(T\), when occupying a volume \(V\) will be
The molecular mass of nitrogen gas \((\text{N}_2)\) is \(28\text{ g/mol}\). The number of moles is \(n = \frac{14}{28} = 0.5\). Thus, using \(PV = nRT\), we get \(PV = \frac{1}{2}RT\).
Question 303:
easyThe unit of emissive power is
Emissive power is defined as the thermal energy emitted per unit area per unit time, so its unit is \(\text{J m}^{-2}\text{ s}^{-1}\) (or \(\text{W m}^{-2}\)).
Question 304:
moderateConsider a sample of \(n\) moles of rigid diatomic gas. Match the columns and tick the correct option (symbols have their usual meanings):
| 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 of \(n\) moles is \(\frac{3}{2}nRT\) (R). Rotational KE is \(nRT\) (Q). KE per mole of diatomic gas is \(\frac{5}{2}RT\) (S). KE per molecule is \(\frac{5}{2}k_B T\) (P). Thus, (A)-(R), (B)-(Q), (C)-(S), (D)-(P).
Question 305:
easyTwo rods one made of material A and other made of material B of same length and same cross-sectional area are joined together. If thermal conductivity of material A is \(K_1\) while that of material B is \(K_2\) and the free end of rod made of material A is maintained at \(T_1\) while that of the rod of material B is maintained at \(T_2\), then the temperature of junction is (Where \(T_1 > T_2\))
Under steady state, the rate of heat flow is the same through both rods: \(\frac{K_1 A(T_1 - T_j)}{L} = \frac{K_2 A(T_j - T_2)}{L}\). Solving for \(T_j\) gives \(T_j = \frac{K_1 T_1 + K_2 T_2}{K_1 + K_2}\).
Question 306:
easyConsider the following statements out of which one is labelled as assertion and other as reason.
Assertion: The internal energy of an ideal monoatomic gas enclosed in a container does not change when there is no change in temperature.
Reason: Internal energy of a gaseous system is path function.
Internal energy of an ideal gas depends only on its temperature, so the Assertion is true. However, internal energy is a state function, not a path function, so the Reason is false.
Question 307:
easyA flask contains hydrogen and oxygen gas in the ratio of 3 : 1 by mass at temperature 27°C. The ratio of average translational kinetic energy per molecule of hydrogen and oxygen respectively is
The average translational kinetic energy per molecule of any gas is given by \(\frac{3}{2} k_B T\). Since both gases are at the same temperature, the ratio of their translational kinetic energies is 1 : 1.
Question 308:
moderateMatch the columns and tick the correct option. (Symbols have their usual meanings)
\begin{array}{|l|l|}
\hline
\textbf{Column-I} & \textbf{Column-II} \\ \hline
\text{a. } \gamma = \frac{5}{3} & \text{(i) Diatomic gas} \\ \hline
\text{b. } C_v = \frac{5}{2}R & \text{(ii) Triatomic non-linear gas} \\ \hline
\text{c. } C_v = 3R & \text{(iii) Monoatomic gas} \\ \hline
\end{array}
Monoatomic gas has \(\gamma = 5/3\) (a-iii). Diatomic gas has \(C_v = 5/2 R\) (b-i). Triatomic non-linear gas has \(C_v = 3R\) (c-ii). Thus, the correct matching is a(iii), b(i), c(ii).
Question 309:
moderateFour moles of helium are mixed with two moles of oxygen. The molar specific heat capacity of the mixture at constant volume is
For helium (monoatomic), \(C_{v1} = \frac{3}{2}R\) and \(n_1 = 4\). For oxygen (diatomic), \(C_{v2} = \frac{5}{2}R\) and \(n_2 = 2\). The mixture molar specific heat is \(C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{4 \times 1.5R + 2 \times 2.5R}{4 + 2} = \frac{11R}{6}\).
Question 310:
easyIn winters, a metal surface feels cooler upon touching than a wooden surface because
Metal is a much better conductor of heat than wood. When touched in winters, heat is rapidly conducted away from our hand to the metal surface, making it feel colder.