Question

Certain mass of a gas occupies 20litre volume at 300k and 2atm pressure.calculate molecules of the gas present in it

Answer 1

Answer:

3.71 * 10²³ molecules.

Explanation:

At first, find the volume of gas at STP by gas laws.

P1 = 2 atm V1 = 20 L. T1 = 300 K

P2 = 1 atm. V2 = ? T2 = 273 K

$\frac{p1 \: \times \: v1}{t1} = \frac{p2 \: \times \: v2}{t2}$

So, V2 = (P1*V1*T2)/(P2*T1)

So, V2 = 36.4 L

Now, we know, 1 mole of any dry gas occupies 22.4 L

So, 22.4 L of the gas gas 6.023 * 10²³ molecules.

So, 36.4 L of the gas has 3.71 * 10²³ molecules.

So, the answer is 3.71 * 10²³ molecules.

Answer 2

Answer:

9.78 × 10²³ molecules

Explanation :

volume gas occupies = 20 litres

at temperature 300K and

2 atm pressure

let,

P1 = 2 atm P2 = 1 atm

T1 = 300K T2 = 273 K

V1 = 20 L V2 = ?? L

p1v1/t1 = p2v2/t2

2×20/300 = 1 × v2/273

v2 = 36.4 L

no. of moles = volume/22.4 L

= 36.4/22.4

= 1.63 mol ~(approx)

no. of moles = no. of molecules/ Avogadro no.

1.63 = no. of molecules/ 6.0221 × 10²³

1.63 × 6 × 10²³ = no. of molecules

So, no. of molecules of the gas present in it =

9.78 × 10²³ (app.)