How Does Doubling Volume and Increasing Temperature Affect Gas Pressure?

AI Thread Summary
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
Messages
9
Reaction score
0
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!
 
Physics news on Phys.org
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"
 
Last edited by a moderator:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top