A sample of neon occupies a volume of 461 mL at 1 atm and 273 K. What will be the volume of the neon when the pressure is reduced to 0.92 atm? (0.501 L)

Question

Simeon

Chemistry

7 March, 05:48

A sample of neon occupies a volume of 461 mL at 1 atm and 273 K. What will be the volume of the neon when the pressure is reduced to 0.92 atm? (0.501 L)

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Answers (1)

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Ainsley Alvarado · Answer 1

0.501 L

Explanation:

To solve this problem we will use Boyle, s Law. According to this law "The volume of given amount of gas is inversely proportional to applied pressure at constant temperature".

V∝ 1/P

V = K/P

VP=K

Here the K is proportionality constant.

so,

P1V1 = P2V2

P = pressure

V = volume

Given dа ta:

P1 = 1 atm

V1 = 461 mL

P2 = 0.92 atm

V2=? (L)

To solve this problem we have to convert the mL into L first.

1 L = 1000 mL

461/1000 = 0.461 L

Now we will put the values in the equation,

P1V1 = P2V2

V2 = P1V1 / P2

V2 = 1 atm * 0.461 L / 0.92 atm

V2 = 0.501 L