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24 February, 22:55

20.352 mL of chlorine under a pressure of 680. mm Hg are

placed in a container

under a pressure of 1210 mm Hg. The temperature remains

constant at 296 K.

What is the volume of the container in liters?

+4
Answers (2)
  1. 25 February, 00:08
    0
    0.01144L or 1.144x10^-2L

    Explanation:

    Data obtained from the question include:

    V1 (initial volume) = 20.352 mL

    P1 (initial pressure) = 680mmHg

    P2 (final pressure) = 1210mmHg

    V2 (final volume) =.?

    Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:

    P1V1 = P2V2

    680 x 20.352 = 1210 x V2

    Divide both side by 1210

    V2 = (680 x 20.352) / 1210

    V2 = 11.44mL

    Now we need to convert 11.44mL to L in order to obtain the desired result. This is illustrated below:

    1000mL = 1 L

    11.44mL = 11.44/1000 = 0.01144L

    Therefore the volume of the container is 0.01144L or 1.144x10^-2L
  2. 25 February, 01:16
    0
    0.011437L

    Explanation:

    In the question, we are told that under a pressure of 680mmHg, chlorine gas occupies a volume of 20.352mL and then the pressure is changed to 1210mmHg at constant temperature.

    Boyle's law states that the volume of a fixed mass of gas is directly proportional to the pressure of the gas at constant temperature.

    Mathematically;

    P1V1=P2V2

    P1 = initial pressure

    V1 = initial volume

    P2 = final pressure

    V2 = final volume

    We will apply Boyle's law to get the new volume.

    From, the relationship P1V1=P2V2

    We make V2 subject of formula

    V2 = (P1V1) / P2

    Given;

    P1=680mmHg

    V1=20.352mL = 20.352/1000L = 0.020352L

    P2=1210mmHg

    V2 = (680*0.020352) / 1210

    V2=0.011437L
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