The equation of state for \(14\text{ g}\) nitrogen gas at a pressure \(P\) and temperature \(T\), when occupying a volume \(V\) will be
1. \(PV = 14RT\)
2. \(PV = \frac{1}{2}RT\)
3. \(PV = RT\)
4. \(PV = 2RT\)
View Answer
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 \(PV = \frac{1}{2}RT\).
The unit of emissive power is
1. \(\text{J m}^{-2}\)
2. \(\text{W s}^{-1}\)
3. \(\text{J m}^{-2}\text{ s}^{-1}\)
4. \(\text{W m}^2\text{ s}^{-1}\)
View Answer
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 \(\text{W m}^{-2}\)).
Consider 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) $\frac{5}{2}RT$ |
1. (A)-(R); (B)-(Q); (C)-(P); (D)-(S)
2. (A)-(Q); (B)-(R); (C)-(S); (D)-(P)
3. (A)-(Q); (B)-(R); (C)-(P); (D)-(S)
4. (A)-(R); (B)-(Q); (C)-(S); (D)-(P)
View Answer
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).
Two 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 \(T_1 > T_2\))
1. \(\frac{T_1 + T_2}{2}\)
2. \(\frac{T_1 K_1 + T_2 K_2}{K_1 + K_2}\)
3. \(\frac{T_1 K_1 - T_2 K_2}{K_1 + K_2}\)
4. \(\frac{T_1 K_1 + T_2 K_2}{K_1 - K_2}\)
View Answer
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 \(T_j = \frac{K_1 T_1 + K_2 T_2}{K_1 + K_2}\).
Consider 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.
1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.
View Answer
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.
Match 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}
1. a(iii), b(ii), c(i)
2. a(iii), b(i), c(ii)
3. a(i), b(ii), c(iii)
4. a(i), b(iii), c(ii)
View Answer
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).
Four moles of helium are mixed with two moles of oxygen. The molar specific heat capacity of the mixture at constant volume is
1. \(\frac{13R}{6}\)
2. \(\frac{11R}{6}\)
3. \(\frac{11R}{2}\)
4. \(\frac{13R}{3}\)
View Answer
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 \(C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{4 \times 1.5R + 2 \times 2.5R}{4 + 2} = \frac{11R}{6}\).
In winters, a metal surface feels cooler upon touching than a wooden surface because
1. Metal has high specific heat capacity as compared to wood
2. Wood has high specific heat capacity as compared to metal
3. Metal has high thermal conductivity
4. Metal has low thermal conductivity
View Answer
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.