Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding work

  1. Dec 19, 2004 #1
    Gas in a container increases its pressure from 206 atm to 3 atm while keepings its volume constant.
    Find the work done by the gas if the volume is 5L. Answer in units of J.
  2. jcsd
  3. Dec 19, 2004 #2


    User Avatar

    Work done equals change in internal energy. If the presure changes then so does the temperature assuming volume doesnt change( which it doesnt) Work out the temps first and then the change in internal energy.
  4. Dec 20, 2004 #3
    Hi girl,
    Work is normally thought of as a transfer of energy due to mechanical action such as a piston compressing the gas or the gas pushing on a piston as it happens in the motor of your car.
    The formula for work is Work = force x distance , but in the case of a gas being compressed or decompressed, it can be shown that the formula above is equivalent to Work = (change in volume) x pressure.
    If your gas in the container changed pressure but the volume stayed constant, then there was no work done (change in volume = 0).
    Another way that energy can be transferred is trough heat, which is defined as a transfer of thermal energy.
    As you can see, the previous post was wrong. There can be a change in internal energy without any work being done. If you take a closed metal or glass container with a gas in it and put it on the stove, it will warmup. Then the gas pressure will increase but because the volume stays the same, there is no work done.
  5. Dec 20, 2004 #4
    If what I wrote is a little confusing, the answer to your homework problem is:
    W = P x delta V
    W = P[tmospheres] x 0L
    W= 0 Liter-Atmosphere = 0J (zero Joules)

    The above formula considers constant pressure. In that case, as the change in volume is zero, it does not matter if the pressure changes or not.
    If both the volume and pressure were changing then the formula would be:
    W = Integral [P(V)dv]

    If your problem implied doing calculations you woud have to convert from liter-Atmosphere to Joules. There may be a table of conversions somewhere in your textbook.
    Otherwise, convert from Liters to cubic meters and from Atmospheres to Pascals. Using cubic meteres and Pascals you get Joules directly.
  6. Jan 2, 2005 #5


    User Avatar

    Zero change in volume implies zero work. Period.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook