21. (JEE Main 2017 (Offline) )

The temperature of an open room of volume 30 m³ increases from 17^oC to 27^oC due to the sunshine. The atmospheric pressure in the room remains 1 $\times$ 10⁵ Pa. If N_i and N_f are the number of molecules in the room before and after heating, then N_f – N_i will be :

(A) - 1.61 $\times$ 10²³

(B) 1.38 $\times$ 10²³

(C) 2.5 $\times$ 10²⁵

(D) - 2.5 $\times$ 10²⁵

Correct answer is (D)

Given: Initial temperature T_i= 17 + 273 = 290 K

Final temperature T_f = 27 + 273 = 300 K

Atmospheric pressure, P₀ = 1 × 10⁵ Pa

Volume of room, V₀ = 30 m³

Difference in number of molecules, N_f – N_i = ?

We know PV = nRT = $\frac{N}{N_{A}}$ RT

The number of molecules

N = $\frac{P V N_{A}}{R T}$

$∴$ N_f – N_i = $\frac{P_{0} V_{0} N_{A}}{R} (\frac{1}{T_{f}} - \frac{1}{T_{i}})$

= $\frac{1 \times 10^{5} \times 30 \times 6.023 \times 10^{23}}{8.314} (\frac{1}{300} - \frac{1}{290})$

= – 2.5 $\times$ 10²⁵

22. (JEE Main 2016 (Online) 10th April Morning Slot )

Which of the following shows the correct relationship between the pressure ‘P’ and density $ρ$ of an ideal gas at constant temperature ?

(A) JEE Main 2016 (Online) 10th April Morning Slot Physics - Heat and Thermodynamics Question 273 English Option 1

(B) JEE Main 2016 (Online) 10th April Morning Slot Physics - Heat and Thermodynamics Question 273 English Option 2

(C) JEE Main 2016 (Online) 10th April Morning Slot Physics - Heat and Thermodynamics Question 273 English Option 3

(D) JEE Main 2016 (Online) 10th April Morning Slot Physics - Heat and Thermodynamics Question 273 English Option 4

Correct answer is (D)

We know, ideal gas equation,

PV = nRT

Here T = constant.

$∴$    PV = constant

$\Rightarrow$    P $\frac{m}{ρ}$ = constant

$\Rightarrow$    P $\propto$ $ρ$

23. (JEE Main 2016 (Offline) )

$^{'} n^{'}$ moles of an ideal gas undergoes a process $A$ $\to$ $B$ as shown in the figure. The maximum temperature of the gas during the process will be :

JEE Main 2016 (Offline) Physics - Heat and Thermodynamics Question 297 English

(A) $\frac{9 P_{0} V_{0}}{2 n R}$

(B) $\frac{9 P_{0} V_{0}}{n R}$

(C) $\frac{9 P_{0} V_{0}}{4 n R}$

(D) $\frac{3 P_{0} V_{0}}{2 n R}$

Correct answer is (C)

JEE Main 2016 (Offline) Physics - Heat and Thermodynamics Question 297 English Explanation

The equation for the line is

$P = \frac{- P_{0}}{V_{0}} V + 3 P$

[slope $=$ $\frac{- P_{0}}{V_{0}}, c = 3 P_{0}$ ]

$P V_{0} + P_{0} V = 3 P_{0} V_{0} . . . (i)$

But $p v = n R T$

$∴$ $p = \frac{n R T}{v} . . . (i i)$

From $(i)$ & $(i i)$ $\frac{n R T}{v} V_{0} + P_{0} V = 3 P_{0} V_{0}$

$∴$ $n R T V_{0} + P_{0} V^{2} = 3 P_{0} V_{0} . . . (i i i)$

For temperature to be maximum $\frac{d T}{d v} = 0$

Differentiating e.q. $(i i i)$ by $^{'} v^{'}$ we get

$n R V_{0} \frac{d T}{d v} + P_{0} (2 v) = 3 P_{0} V_{0}$

$∴$ $n R V_{0} \frac{d T}{d v} = 3 P_{0} V_{0} - 2 P_{0} V$

$\frac{d T}{d v} = \frac{3 P_{0} V_{0} - 2 P_{0} V}{n R V_{0}} = 0$

$V = \frac{3 V_{0}}{2}$

$∴$ $p = \frac{3 P_{0}}{2}$ [From $(i)$ ]

$∴$ $T_{max} = \frac{9 P_{0} V_{0}}{4 n R}$ [ From $(i i)$ ]

Kinetic Theory of Gases