Question

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 1

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.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 0.988 atm

$P_2$ = final pressure = ?

$V_1$ = initial volume = 1.00 L

$V_2$ = final volume = 2.00 L

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

$0.988atm\times 1.00L=P_2\times 2.00L$

$P_2=0.494atm$

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