Question

If the pressure of the gas in a 2.31 L balloon is 0.12 ATM and the volume increases to 7.14 L what will be the final pressure of the air within the balloon? Assume the temperature remains constant

Answer 1

Answer: 0.04 atm

Step-by-step explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

`P\propto (1)/(V)` (At constant temperature and number of moles)

`P_1V_1=P_2V_2`

where,

`P_1` = initial pressure of gas = 0.12 atm

`P_2` = final pressure of gas = ?

`V_1` = initial volume of gas =
`2.31 L`

`V_2` = final volume of gas = 7.14 L

`0.12* 2.31=P_2* 7.14`

`P_2=0.04atm`

Therefore, the final pressure of the air within the balloon is 0.04 atm.

Answer 2

We can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

P2 = P1 x V1 / V2

P2 = 0.12 x 2.31 / 7.14

P2 = 0.04 atm