 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping • Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details

Solved Examples of Kinetic Theory of Gases

Problem 1 (JEE Main):-

Calculate the number of molecules in 210-6 m3 of a perfect gas at 27°C and at a pressure of 0.01 m of mercury. Mean K.E of a molecule at 27°C = 410-11 J and g = 9.8 ms-2.

Solution:-

P = (1/3) (M/V) C2 or PV = (1/3) MC2

But M = (m) (n)

Where ‘m’ is mass of one molecule and n is the number of molecules.

PV = (1/3) (m) C2

n = 3PV/mC2 = [3/2 PV] / [(1/2) mC2]

Here, P = 0.01 m of mercury = 0.01136009.8Nm-2

V =  210-6 m3 = 1332.8 Nm-2

½ mc2 = 410-11J

Thus from the above observation we conclude that, the number of molecules would be 9.996107.

_____________________________________________________________________________________ Problem 2:-

You are throwing a birthday party and decide to fill the room with helium balloons. You also want to have a few larger balloons to put at the door. The smaller balloons are filled occupy 0.240 m3 when the pressure inside them is 0.038 atm and the temperature of the room is 70° F. What pressure should you fill the larger balloons to so that they occupy 0.400 m3?

Solution:-

The temperature of the room is assumed to be held constant, so it is extraneous information. Since you are dealing with volume and pressure, you would use Boyle's Law.

In accordance to data,

P1= 0.038 atm, V1= 0.240 m3

V2=0.400 m3

We have to find out V2.

Substitute the vale of P1, V1 and V2 in equation P1V1=P2V2

P1V1=P2V2

(0.038 atm)(0.240 m3) = P2(0.400 m3)

0.00912 = 0.400 P2             (units cancel out so that pressure will be in atm)

P2 = 0.0228 atm

From the above observation we conclude that, the pressure would be 0.0228 atm.

________________________________________________________________________________________

Problem 3:-

Find the RMS speed of a sample of neon gas at 80° F.

Solution:-

First convert Fahrenheit to Celsius.

°C = (°F- 32)/1.8

= (80 -32)/1.8

= 48/1.8

= 26.6

Convert Celsius to Kelvin.

K = °C + 273.15

= 26.6 + 273.15

= 299.75 K

Substitute the known information into the equation for RMS speed and solv, we get,.

vrms = √3RT/Mx

= √3(8.3144) (299.75)/ (20.179)

= 19.2 m/s

From the above observation we conclude that, the RMS speed of a sample of neon gas at 80° F would be 19.2 m/s.  ### Course Features

• 728 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution 