nattycat1105
contestada

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?​

Respuesta :

Eduard22sly

Answer:

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

Answer:

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

Otras preguntas