A gas occupies 4.00 L at 2.00 atm pressure. At what pressure would the volume be 6.00 L if the temperature remains constant?

Respuesta :

P1V1= P2V2 
P1= 2
V1=4
V2=6
answere= (8/6)

Answer:

The final pressure would be 1.33 atm

Explanation:

Given:

Initial volume of the gas, V1 = 4.00 L

Initial pressure, P1 = 2.00 atm

Final volume, V2 = 6.00 L

To determine:

The final pressure P2 of the gas

Explanation:

Based on the ideal gas equation:

[tex]PV = nRT[/tex]

where: n = number of moles,

R = gas constant and T = temperature

At constant Temperature the gas law reduces to,

[tex]PV = constant\\\\P1V1 = P2V2\\\\P2 = \frac{P1V1}{V2} = \frac{2.00atm*4.00L}{6.00L} =1.33\ atm[/tex]

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