16. (JEE Main 2020 (Online) 4th September Morning Slot )

Match the $\frac{C_{P}}{C_{V}}$ ratio for ideal gases with different type of molecules :

Molecule Type	C_P/C_V
(A) Monatomic	(I) 7/5
(B) Diatomic rigid molecules	(II) 9/7
(C) Diatomic non-rigid molecules	(III) 4/3
(D) Triatomic rigid molecules	(IV) 5/3

(A) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

(B) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

(C) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

(D) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Correct answer is (C)

$γ = C_{p} / C_{v}$

$γ_{A} = 1 + \frac{2}{3} = 5 / 3$

$γ_{B} = 1 + \frac{2}{5} = 7 / 5$

$γ_{C} = 1 + \frac{2}{7} = 9 / 7$

$γ_{D} = 1 + \frac{2}{6} = 4 / 3$

17. (JEE Main 2020 (Online) 3rd September Evening Slot )

To raise the temperature of a certain mass of gas by 50^oC at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by 100^oC at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

(A) 6

(B) 7

(C) 5

(D) 3

Correct answer is (A)

$160 = n C_{p} 50$ ....(i)

$240 = n C_{v} 100$ ....(ii)

Dividing (i) by (ii), we get

$\frac{160}{240} = \frac{C_{p}}{C_{v}} \times \frac{1}{2}$

$\Rightarrow$ $\frac{C_{p}}{C_{v}}$ = $\frac{4}{3}$

We know,

$γ = \frac{C_{p}}{C_{v}} = 1 + \frac{2}{f} = \frac{4}{3}$

$\Rightarrow f = 6$

18. (JEE Main 2020 (Online) 9th January Morning Slot )

Consider two ideal diatomic gases A and B at some temperature T. Molecules of the gas A are rigid, and have a mass m. Molecules of the gas B have an additional vibrational mode, and have a mass $\frac{m}{4}$ . The ratio of the specific heats ( $C_{V}^{A}$ and $C_{V}^{B}$ ) of gas A and B, respectively is :

(A) 7 : 9

(B) 5 : 7

(C) 3 : 5

(D) 5 : 9

Correct answer is (B)

Degree of freedom of a diatomic molecule if vibration is absent = 5

Degree of freedom of a diatomic molecule if vibration is present = 7

$∴$ $C_{V}^{A}$ = $\frac{5}{2} R$

and $C_{V}^{B}$ = $\frac{7}{2} R$

$∴$ $\frac{C_{V}^{A}}{C_{V}^{B}} = \frac{5}{7}$

19. (JEE Main 2020 (Online) 8th January Evening Slot )

Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its C_P/C_V value will be :

(A) 23/15

(B) 67/45

(C) 40/27

(D) 19/13

Correct answer is (D)

$\frac{C_{P}}{C_{V}} = \frac{n_{1} C_{P_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}$

= $\frac{n \times \frac{5 R}{2} + 2 n \times \frac{7 R}{2}}{n \times \frac{3 R}{2} + 2 n \times \frac{5 R}{2}}$

= $\frac{5 + 14}{3 + 10}$

= $\frac{19}{13}$

20. (JEE Main 2020 (Online) 7th January Morning Slot )

Two moles of an ideal gas with $\frac{C_{P}}{C_{V}} = \frac{5}{3}$ are mixed with 3 moles of another ideal gas with $\frac{C_{P}}{C_{V}} = \frac{4}{3}$ . The value of $\frac{C_{P}}{C_{V}}$ for the mixture is :

(A) 1.50

(B) 1.45

(C) 1.47

(D) 1.42

Correct answer is (D)

C_p = $\frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} + n_{2}}$

C_v = $\frac{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}{n_{1} + n_{2}}$

$γ$ _mix = $\frac{C_{p}}{C_{v}}$ = $\frac{2 \times \frac{5}{2} R + 3 \times \frac{8}{2} R}{2 \times \frac{3}{2} R + 3 \times \frac{6}{2} R}$

= $\frac{5 + 12}{3 + 9}$ = 1.42

Kinetic Theory of Gases