How can pressure and volume affect the work, energy, and heat of helium gas?

In summary, a compression work problem is a type of physics problem that involves calculating the work done on a system when it is compressed. This is typically measured in joules and can be used to determine the change in energy of the system. The formula for calculating work in a compression work problem is W = F x d, where W is work, F is the force applied, and d is the distance the system is compressed. Negative work indicates energy being transferred out of the system, while positive work indicates energy being transferred into the system. Real-life examples of compression work problems include pressing a spring, squeezing a sponge, or compressing a gas in a container. The compression ratio, which is the ratio of the volume of the system before compression
  • #1
pentazoid
146
0

Homework Statement


Imagine some helium in a cylinder with initial volume of 1 liter and an initial pressure of 1 atm. Somehow the helium is made to expand to a final volume of 3 liters, in such a way that its pressure rises in diect proportion to its volume.

a) calculate the work done on the gas during this process , assuming that there are no "other" types of work being done.

b) Calculate the change in the helium's energy content during this process

c) calculate the amount of heat added to or removed from the helium during this process

d) Describe what you might do to cause the pressure o rise as the helium expands


Homework Equations



W=-integral(from V_i to V_f) P(V) dV (quasistatic)

W=-P*delta(V)

delta(U)= W+Q

delta(U)=f/2*NkT

The Attempt at a Solution



a) P=m*V+ b , where m is the slope and b is the y-int.

m=delta(P)/delta(V)= (3-1)/((3-1)=1 and b=0

P=V

W=integral (from 1 atm to 3 atm) (V dV)


b) delta(U)=f/2*NkT=(f/2)*PV

c)delta(U)=Q+W==> Q=delta(U)-W

d) some source of external heat such as a flame is probably causing the balloon to expand

Did I do everything correctly?
 
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  • #2
P is directly proportional to V so

P = kV

where k is a constant.

Use this value for P in the work equation and integrate.
 
  • #3
chrisk said:
P is directly proportional to V so

P = kV

where k is a constant.

Use this value for P in the work equation and integrate.

Are my remaining solutions to the rest of my problems correct?
 
  • #4
pentazoid said:
Are my remaining solutions to the rest of my problems correct?

I thought k would just equal 1 since when you find the equation for P, P=V
 
  • #5
The value k is a proportionality constant. One must know initial and final values of V and P to determine k. Since this was not given, the value k is unknown. Your answers will be in terms of k.
 
  • #6
chrisk said:
The value k is a proportionality constant. One must know initial and final values of V and P to determine k. Since this was not given, the value k is unknown. Your answers will be in terms of k.

but the initial and final values of P and V were given in the problem. here is a quote from the question. :"Some how the helium is made to expand to a final volume of 3 liters , in such a way that its pressure rises in direct proportion to its volume". So if my volume rises to 3 liters, then my pressure should rise to 3 atm.
 
  • #7
Please indicate what the final pressure is according to problem. There is no final pressure stated from what is given, only that the pressure varies directly with the volume.
 
  • #8
chrisk said:
Please indicate what the final pressure is according to problem. There is no final pressure stated from what is given, only that the pressure varies directly with the volume.

The problem doesn't state what the final pressure is. It just says that pressure varies in direct proportion to its volume. The problem says that there is an initial volume of 1 liter and an initial pressure of 1 atm. Since the pressure rises in direct proportion with its volume, if the final volume is 3 liters, doesn 't that mean the final pressure is 3 atm.
 
  • #9
No, it does not imply the final pressure is 3 atm. So, use k instead. Recall, k is the slope of the line in this case and two points are required to determine the slope of a line; you do not have the P coordinate of the second point.
 

Related to How can pressure and volume affect the work, energy, and heat of helium gas?

What is a compression work problem?

A compression work problem is a type of physics problem that involves calculating the work done on a system when it is compressed. This is typically measured in joules, and can be used to determine the change in energy of the system.

How do you calculate the work done in a compression work problem?

The formula for calculating work in a compression work problem is W = F x d, where W is work, F is the force applied, and d is the distance the system is compressed. The unit of work is joules (J).

What is the difference between negative and positive work in a compression work problem?

Negative work in a compression work problem indicates that energy is being transferred out of the system, while positive work indicates that energy is being transferred into the system. In other words, negative work decreases the energy of the system, while positive work increases it.

What are some real-life examples of compression work problems?

Some examples of compression work problems in real life include pressing a spring, squeezing a sponge, or compressing a gas in a container. In all of these situations, work is done on the system to compress it, resulting in a change in energy.

How does the compression ratio affect the work done in a compression work problem?

The compression ratio, which is the ratio of the volume of the system before compression to the volume after compression, directly affects the work done. The greater the compression ratio, the more work is needed to compress the system, resulting in a larger change in energy.

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