How Does Doubling Volume and Increasing Temperature Affect Gas Pressure?

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Doubling the volume of an ideal gas and increasing the temperature to 50 degrees Celsius affects its gauge pressure, which can be calculated using the Ideal Gas Law (PV = nRT). At standard temperature and pressure (STP), the initial conditions are 6.00 L at 0 degrees Celsius and 1 atm. To find the new pressure after the changes, the temperature must be converted to Kelvin, and the new volume must be considered. The Ideal Gas Law provides the necessary framework to solve for the new pressure. Understanding these relationships is essential for accurately determining gas behavior under varying conditions.
Shakerhood
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An Ideal Gas occupies a Volume of 6.00 L at STP. What is its Gauge pressure (in kPa) if the Volume is doubled and the Temperature is increased to 50.0 Degrees Celcius. I don't know where to begine as the question doesn't have a starting Temp, just that it was increased to 50 Degrees Celcius. Any help is greatly appreciated!
 
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STP stands for standard temperature and pressure. it is 0 degrees C and 1 atm.
 
Do you have an equation for me that might help?
 
The Ideal Gas Law should cover this.
 
whozum said:
The Ideal Gas Law should cover this.

I have a really Sorry Book, I will try to find that in there. Thanks!
 
Shakerhood said:
I have a really Sorry Book, I will try to find that in there. Thanks!

The Ideal Gas law says in an Ideal Gas in a closed system the following relation always holds constant,

PV = nRT

P = Pressure
V = Volume
n = number of moles
R = The gas constant
T = Temperature (KELVIN)

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/idegas.html"
 
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