A sealed syringe contains 10 × 10−6 m3 of air at 1 × 105 Pa. The plunger is pushed until the volume of trapped air is 4 × 10−6 m3. If there

Question

A sealed syringe contains 10 × 10−6 m3 of air at 1 × 105 Pa. The plunger is pushed until the volume of trapped air is 4 × 10−6 m3. If there is no change in temperature, what is the new pressure of the gas?

in progress 0
King 4 years 2021-07-28T08:42:47+00:00 1 Answers 94 views 0

Answers ( )

  1. minhkhue
    1
    2021-07-28T08:44:27+00:00
    Reply

    Answer:P_2=2.5\times 10^5\ Pa

    Explanation:

    Given

    Initial volume V_1=10\times 10^{-6}\ m^3

    Initial Pressure P_1=10^5\ Pa

    Final trapped air V_2=4\times 10^{-6}\ m^3

    If there is no change in temperature then We can write

    P_1V_1=P_2V_2\quad \text{(From ideal gas equation)}

    P_2=\dfrac{P_1V_1}{V_2}

    P_2=10^5\times \frac{10\times 10^{-6}}{4\times 10^{-6}}

    P_2=2.5\times 10^5\ Pa

Leave an answer

Browse

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