1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: Ideal Gases/Pressure

  1. Jul 10, 2009 #1
    1. The problem statement, all variables and given/known data
    An ideal gas that occupies 1.8 m3 at a pressure of 1.0 multiplied by 105 Pa and a temperature of 27°C is compressed to a volume of 0.80 m3 and heated to a temperature of 227°C. What is the new pressure?

    2. Relevant equations

    3. The attempt at a solution
    Not really sure what to do for this one, not really even sure this is the formula I'd want to use
  2. jcsd
  3. Jul 10, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    That's the correct formula, except that 'nR' is constant so it simply becomes
    PV/T (before) = PV/T (after)

    Note that you need to use absolute (kelvin) temperatures
  4. Jul 10, 2009 #3
    so would it be (1.80)(1.0x10^5)/27=P(0.80)/227?
  5. Jul 10, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    Except you need absolute temperatures
  6. Jul 10, 2009 #5
    This problem seems to involve a simple IDEAL GAS. Remember, Ideal gases involves molecules of gas as POINT particles that do not involve electric forces nor volumes that molecules can fill. Using the PV = nRT seems like the correct choice. DO NOT FORGET that units are key for these first steps into studying ThermoDynamics. R = 8.314 Joules / (Mole x Kelvin) this R value is in SI units. I usually convert everything to SI units before moving forward.

    R can also = .0821 (L x Atm) / (Mole x kelvin)

    You mentioned Pascal. 1 pascal = 1 Newton/Meter squared.

    the atkins textbook plots Pressure, Volume, and Temperature on the X, Y, Z axis. try to plot a couple of Pressure and Temperature points. Also, try plotting Volume and Temperature points. try to think about how temperature, volume of gas, and pressure of gas are related to each other when their Kinetic Energies are relatively High like in gases.
  7. Jul 10, 2009 #6
    mgb_phys is underlining something very key into picking out the right values to use in your problem solving. Don't just blindly accept the given value as the value to use to plug into your formula. Instead, try to think about how the temperature was taken because 227 degrees Celsius can refer to temperature of the environment plus the temperature of the isolated gas system. Just like Gauge pressure is different from total pressure due to the environment, absolute temperature can differ from total temperature.
  8. Jul 11, 2009 #7
    so would it be (1.80)(1.0x10^5)/300=P(0.80)/500?

    somehow I don't think so
  9. Jul 11, 2009 #8
    It's correct.
  10. Jul 11, 2009 #9
    oh yeah...I put a parenthese in the wrong place when I put in my calculator...opps. Thanks
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook