Chemistry, asked by Dorji123, 10 months ago

A sample of neon occupies a volume of 461ml at STP. What will be the volume of the neon when the pressure is reduced to 93.3kpa

Answers

Answered by RpRai1753
3

Answer:

Heya.......!!!!!

It's a question of Boyle's Law .

Given that :

=> P1 = 101.3 kPa

=> P2 = 93.3 kPa

=> V1 = 461 mL

=> V2 = ?

By equation of Boyle's Law

P1V1 = P2V2 .

=>> V2 = P1V1/P2

V2 = 101.3 x 461 /93.3

➡ V2 = 501 ml Neon .

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Answered by nraj73523paiks1
0

Answer:

Do it like this=>

P1V1/T1=P2V2/T2 where P1,P2 are the pressure and V1,V2 are volumes

and as temperature is same then T1 cancels T2

so,

P1V1=P2V2

1atm×461ml=xml×93.3kpa

101.3kpa×461ml=93.3kpa×xml(1atm=101.3 kpa)

x=nearly 500.53 ml

Explanation:

HOPE IT HELPS PLZ MARK AS BRAINLEST

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