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A 5.00-L tank contains helium gas at 1.50 atm. What is the pressure of the gas when the volume increased to 4.5atm?

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Answer:

the pressure of the gas is 1.67 atm when the volume increased to 4.5 L.

Step-by-step explanation:

Assuming the temperature and the amount of gas remain constant (i.e., the process is isobaric):

Using Boyle's law, we can relate the initial pressure (P1) and volume (V1) to the final pressure (P2) and volume (V2):

P1V1 = P2V2

Plugging in the given values:

P1 = 1.50 atm

V1 = 5.00 L

V2 = 4.5 L

Solving for P2:

P2 = (P1V1)/V2 = (1.50 atm x 5.00 L)/4.5 L = 1.67 atm

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