Question

What is the velocity of 1.4 moles of bromine in a 3.5 L container at 4.35 atm?

Answer 1

Step-by-step explanation:

Let the temperature at which gas is moving be T

Pressure exerted by the gas , P = 4.35 atm

Volume occupied by the gas ,V = 3.5 l

Moles of bromine gas,n = 1.4 moles

Consider the bromine gas is behaving ideal .

`PV=nRT`

`4.35 atm* 3.5 L=1.4mol* 0.0820 L atm/K mol* T`

`T=(4.35 atm* 3.5 L)/(1.4mol* 0.0820 L atm/K mol)=132.62 K`

Mass of bromine gas,M = Moles × Molar mass = 1.4 mole × 160 g/mol = 224 g = 0.224 kg

Value of R = 8.314 J/mol K

Root mean square velocity of the bromine gas:

`\mu_(rms)=\sqrt{(3RT)/(M)}=\sqrt{(3* 8.314 J/mol K* 132.62 K)/(0.224 kg)}=121.51 m/s`

Average velocity of the bromine gas:

`v_(avg)=\sqrt{(8RT)/(\piM)}=\sqrt{(8* 8.314 J/mol K* 132.62 K)/(3.14* 0.224 kg)}=111.98 m/s`

Most probable velocity:

`v_(m)=\sqrt{(2RT)/(M)}=\sqrt{(2* 8.314 J/mol K* 132.62 K)/(0.224 kg)}=99.22 m/s`