The pressure of a sample of helium in a 1.00-L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00-L container?

Answer :

Answer : The new pressure if the sample placed in a 2.00-L container is, 0.494 atm.

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]

or,

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure = 0.988 atm

[tex]P_2[/tex] = final pressure = ?

[tex]V_1[/tex] = initial volume = 1.00 L

[tex]V_2[/tex] = final volume = 2.00 L

Now put all the given values in the above equation, we get:

[tex]0.988atm\times 1.00L=P_2\times 2.00L[/tex]

[tex]P_2=0.494atm[/tex]

Thus, the new pressure if the sample placed in a 2.00-L container is, 0.494 atm.

The new pressure if the sample is placed in a 2.00-L container is 0.494 atm.

Boyle's Law:

It means the pressure of the gas should be inversely proportional with respect to the volume of the gas at the same temperature also the no of moles.

P1V1 = P2V2

Here

P1 = initial pressure = 0.988 atm

P2 = final pressure = ?

V1 = initial volume = 1.00 L

V2 = final volume = 2.00 L

So,

the new pressure should be

0.988 * 1.00L = P2 * 2.00l

So, P2 is 0.494 atm

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