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

Question

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?

in progress 0
Euphemia 5 months 2021-08-27T20:29:57+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-08-27T20:31:40+00:00

    Answer:

    The final value of the pressure is 0.89 atm

    Explanation:

    Considering the formula for constant temperature

    \frac{P1 V1}{nR} = \frac{P2 V2}{nR}\\

    P1 V1 = P2 V2

    P2 = \frac{P1 V1}{V2} = \frac{0.11 * 3.50}{0.430} atm = 0.89 atm

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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )