A ballon filled with helium contains 85L and 1.2 atm. the pressure is reduced to 0.22 atm at constant temperature. what is the new volume? given info: equation. work: answer

463.64L

Explanation:

The following data were obtained from the question:

V1 (original volume) = 85L

P1 (original pressure) = 1.2 atm

P2 (new pressure) = 0.22 atm

V2 (new volume) =?

Since the temperature is constant, the gas is obeying Boyle's law.

Using the Boyle's law equation P1V1 = P2V2, the new volume of the gas can obtain as follow:

P1V1 = P2V2

1.2 x 85 = 0.22 x V2

Divide both side by 0.22

V2 = (1.2 x 85)/0.22

V2 = 463.64L

Therefore, the new volume is 463.64L

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"