Match list-I with list-II about effect of various factors on speed of sound in a gas.
\[\begin{array}{|l|l|} \hline
\text{List-I (Factor)} & \text{List-II (Speed)} \\ \hline
\text{a. With increase in temperature} & \text{(ii) Increases} \\ \hline
\text{b. With increase in molecular weight of gas (temp const)} & \text{(iii) Decreases} \\ \hline
\text{c. With increase in pressure at constant temperature} & \text{(i) Same} \\ \hline
\end{array}\]
1. a(ii), b(i), c(iii)
2. a(ii), b(iii), c(i)
3. a(iii), b(ii), c(i)
4. a(i), b(ii), c(ii)
View Answer
Speed of sound is \(v = \sqrt{\frac{\gamma RT}{M}}\). Temperature increase increases speed, molecular weight increase decreases speed, and pressure has no effect at constant temperature.
In a thermodynamic process, pressure (in Pa) varies as, \(P = a + bV\) [where \(V\) represents volume in \(\text{m}^3\), \(a\) and \(b\) are constants]. The work done by gas during expansion from \(2 \text{m}^3\) to \(3 \text{m}^3\) will be
1. \(\left(\frac{3a}{2} + b \right) \text{J}\)
2. \(\left(a + \frac{5b}{2}\right) \text{J}\)
3. \(\left(b + \frac{5a}{2}\right) \text{J}\)
4. Zero
View Answer
Work done \(W = \int_{V_1}^{V_2} P dV = \int_2^3 (a+bV) dV = \left[aV + \frac{bV^2}{2}\right]_2^3 = a(3-2) + \frac{b}{2}(9-4) = a + \frac{5b}{2}\).
Four particles have velocities 1, 0, 2 and 3 m/s. The root mean square velocity of the particles is
1. \(\sqrt{3.5} \text{m/s}\)
2. \(\sqrt{5.3} \text{m/s}\)
3. \(\sqrt{2.8} \text{m/s}\)
4. \(\sqrt{4.7} \text{m/s}\)
View Answer
The RMS velocity is given by \[v_{\text{rms}} = \sqrt{\frac{v_1^2 + v_2^2 + v_3^2 + v_4^2}{4}}\]. Substituting the values, \[v_{text{rms}} = \sqrt{\frac{1^2 + 0^2 + 2^2 + 3^2}{4}} = \sqrt{\frac{14}{4}} = \sqrt{3.5} \text{m/s}\].
A gas mixture consists of \(2\) moles of \(\text{O}_2\) and \(4\) moles of \(\text{He}\) at temperature \(T\). Neglecting all vibrational modes, total internal energy of the system is
1. \(11\text{ }RT\)
2. \(13 RT\)
3. \(15 RT\)
4. \(9 RT\)
View Answer
Internal energy \(U = n_1 \frac{f_1}{2} RT + n_2 \frac{f_2}{2} RT\). For diatomic \(\text{O}_2\), \(f_1 = 5\), and for monoatomic \(\text{He}\), \(f_2 = 3\). Thus, \(U = 2 \left(\frac{5}{2}\right) RT + 4 \left(\frac{3}{2}\right) RT = 5RT + 6RT = 11RT\).
| 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)-(S); (D)-(P)
2. (A)-(Q); (B)-(R); (C)-(S); (D)-(P)
3. (A)-(Q); (B)-(R); (C)-(P); (D)-(S)
4. (A)-(R); (B)-(Q); (C)-(P); (D)-(S)
View Answer
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 \(\frac{5}{2}K_B T\).
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).
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 \(C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{4 \times 1.5R + 2 \times 2.5R}{4 + 2} = \frac{11R}{6}\).