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**PROBLEM 1 (JEE MAIN):-****Calculate the number of molecules in 2 x 10 ^{-6} m^{3} of a perfect gas at 27Â°C and at a pressure of 0.01 m of mercury. Mean K.E of a molecule at 27Â°C = 4 x 10^{-11} J and g = 9.8 ms^{-2}.**

P = (1/3) (M/V) C

But M = (m) (n)

Where â€˜mâ€™ is mass of one molecule and n is the number of molecules.

PV = (1/3) (m) C

n = 3PV/mC

Here, P = 0.01 m of mercury = 0.01 x 13600 x 9.8Nm

V = 2 x 10

Â½ mc

Thus from the above observation we conclude that, the number of molecules would be 9.996 x 10

**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 70Â° F. What pressure should you fill the larger balloons to so that they occupy 0.400 m ^{3}?**

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,

P

V

We have to find out V

Substitute the vale of P

P

(0.038 atm)(0.240 m

0.00912 = 0.400 P

P

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 80Â° F.****Solution:-**

First convert Fahrenheit to Celsius.

Â°C = (Â°F- 32)/1.8

= (80 -32)/1.8

= 48/1.8

= 26.6

Convert Celsius to Kelvin.

K = Â°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 = âˆš3RT/M_{x }

= âˆš3(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 80Â° F would be 19.2 m/s.

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