promotion image of download ymail app
Promoted
Anonymous
Anonymous asked in Science & MathematicsPhysics · 6 months ago

Please help with this problem! Due Soon! It has something to do with VOLUME, TEMP & PRESSURE of 2 chambers. How far does dividing wall move?

Consider a box with two chambers labeled 1 and 2. Each chamber is filled with identical ideal gases that are initially at the same pressure, temperature, and volume; and the vertical wall between the chambers is free to move without transferring any thermal energy or gas between the chambers. Each chamber starts at an initial volume of 2 m^3 and pressure 100 kPa. We then heat the gas of Chamber 1, which causes the dividing wall to move. When the wall comes to rest, Chamber 1 has an absolute temperature that is 50% higher than Chamber 2. Given that the dividing wall has a cross-sectional area of 1 m^2, how far has the wall moved from its original position?   

Could you please include an image of the situation with labels, so the explanation is easier to understand?   

Also, it would be very nice if you could include the principles you used to find your answer.

2 Answers

Relevance
  • Favorite Answer

    I'm going to start with the ideal gas law and spitball this until I figure something out. That's all you can do in some cases.

    P * V = n * R * T

    P * V / T = n * R

    Since both gases are at the same Pressure, Volume, and Temperature initially, then there must be the same number of molecules of gas in each chamber, which means that we can treat n * R as some constant k.

    P * V / T = k

    Initially, P[a1] = P[b1], V[a1] = V[b1] and T[a1] = T[b1]

    Afterwards

    T[a2] = 1.5 * T[b2]

    P[a1] = P[b1] = 100 kPa

    V[a1] = V[b1] = 2 cubic meters

    The cross-sectional area of the wall is 1 square meter. Let's say that it moves y meters. That means that the new volume of chamber 1 is 2 + y cubic meters and the volume of chamber 2 is 2 - y cubic meters.

    V[a2] = 2 + y

    V[b2] = 2 - y

    P[a1] * V[a1] / T[a1] = P[a2] * V[a2] / T[a2]

    P[b1] * V[b1] / T[b1] = P[b2] * V[b2] / T[b2]

    P[a1] * V[a1] / T[a1] = P[b1] * V[b1] / T[b1]

    P[a2] * V[a2] / T[a2] = P[b2] * V[b2] / T[b2]

    P[a2] * (2 + y) / (1.5 * T[b2]) = P[b2] * (2 - y) / T[b2]

    P[a2] * (2 + y) * (2/3) = P[b2] * (2 - y)

    P[a2] * (4 + 2y) = P[b2] * (6 - 3y)

    4 * P[a2] + 2 * P[a2] * y = 6 * P[b2] - 3 * P[b2] * y

    2 * P[a2] * y + 3 * P[b2] * y = 6 * P[b2] - 4 * P[a2]

    y * (2 * P[a2] + 3 * P[b2]) = 6 * P[b2] - 4 * P[a2]

    y = 2 * (3 * P[b2] - 2 * P[a2]) / (3 * P[b2] + 2 * P[a2])

    We know that the Pressure exerted on either side is the same, so P[a2] = P[b2]

    y = 2 * (3 * P[a2] - 2 * P[a2]) / (3 * P[a2] + 2 * P[a2])

    y = 2 * P[a2]) * (3 - 2) / (P[a2]) * (3 + 2))

    y = 2 * (3 - 2) / (3 + 2)

    y = 2 * 1 / 5

    y = 2/5

    It moves over 2/5 of a meter, or 40 cm

    • NCS
      Lv 7
      6 months agoReport

      Nice work. I did not see that this had been answered when I posted.

    • Commenter avatarLogin to reply the answers
  • NCS
    Lv 7
    6 months ago

    Surely you can come up with this simple diagram.

    Initially we have pV = nRT

    where all quantities are the same in both chambers. Throughout the heating, the quantity n*R is constant in both chambers (no gas is exchanged between the chambers. So pV/T is constant.

    We heat the gas in chamber 1. Its temperature increases, so its volume and/or the pressure must also increase in order to keep pV/T constant. In chamber 2, where there is no temperature change, it must be that the pressure increases (to match p1) and the volume decreases (as the dividing wall moves) since the temperature there is constant.

    Since the new pressure in both chambers must be the same as each other, we can write (after the temperature increase)

    p1*V1 / 1.5T = p2*(4m³ - V1) / T

    But p1 = p2, so they cancel, as does the temperature T:

    V1 = 1.5(4m³ - V1)

    2.5*V1 = 6m³

    V1 = 2.4 m³

    which is an increase of 0.4 m³. ◄

    Since the wall has an area of 1 m², it must have moved (toward chamber 2) 0.4 meters.

    If you find this helpful, please award Best Answer!

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.