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A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.120 of air at a pressure of 3.50 .

The piston is slowly pulled out until the volume of the gas is increased to 0.430 . If the temperature remains constant, what is the final value of the pressure?

Respuesta :

Answer:

The final value of the pressure is 0.89 atm

Explanation:

Considering the formula for constant temperature

[tex]\frac{P1 V1}{nR} = \frac{P2 V2}{nR}\\[/tex]

P1 V1 = P2 V2

P2 = [tex]\frac{P1 V1}{V2} = \frac{0.11 * 3.50}{0.430} atm = 0.89 atm[/tex]

This way, considering P1 is in atm P2 is also in atm

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