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Ideal Gases/Pressure

  • Thread starter tag16
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  • #1
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Homework Statement


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?


Homework Equations


PV=nRT


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
 

Answers and Replies

  • #2
mgb_phys
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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
 
  • #3
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so would it be (1.80)(1.0x10^5)/27=P(0.80)/227?
 
  • #4
mgb_phys
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Except you need absolute temperatures
 
  • #5
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Homework Statement


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?


Homework Equations


PV=nRT


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
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.
 
  • #6
69
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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
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.
 
  • #7
97
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so would it be (1.80)(1.0x10^5)/300=P(0.80)/500?

somehow I don't think so
 
  • #8
2,479
98
so would it be (1.80)(1.0x10^5)/300=P(0.80)/500?

somehow I don't think so
It's correct.
 
  • #9
97
0
oh yeah...I put a parenthese in the wrong place when I put in my calculator...opps. Thanks
 

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