Asample 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)

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 data:

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

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