Heat removed from a gas in a cylinder

In summary: Pa and an area of 3.14 x 10^-2 m/s^2. 2093 J of heat is removed from the gas and the task is to find how far the piston has dropped, ignoring mass and friction. To solve this, the isobaric process must be considered, where W = P (Vf-Vi) and the change in distance can be found by multiplying the area by the change in volume. It may also be necessary to use formulas related to specific heat capacities. Additionally, the presence of a monatomic gas may indicate a potential issue with heat capacity.
  • #1
Jared944
10
0

Homework Statement



A piston (with an area of 3.14 x 10^-2 m/s^2) is pushing down on a vertical cylinder (with a pressure of 1.01 x 10^5 Pa) which contains an monotomic ideal gas. 2093 J of heat is removed from the gas.
Ignoring the mass of the piston or friction, find how far the piston has dropped.

Homework Equations



I believe that I will have to use Uf-Ui = Q - w,

but I have already figured that the process would have to be isobaric, whoch is goverend by the equation W = P (Vf-Vi), which can be suited to our needs by using distance multiplied by area for the volume, so
W = P A(change in d)

Im wondering if I am going to have to use formulas related to specific heat capacities, but I am not quite sure. Any ideas?
 
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  • #2
Jared944 said:
A piston (with an area of 3.14 x 10^-2 m/s^2) is pushing down on a vertical cylinder (with a pressure of 1.01 x 10^5 Pa) which contains an monotomic ideal gas. 2093 J of heat is removed from the gas.
Ignoring the mass of the piston or friction, find how far the piston has dropped.

Homework Equations



I believe that I will have to use Uf-Ui = Q - w,

but I have already figured that the process would have to be isobaric, whoch is goverend by the equation W = P (Vf-Vi), which can be suited to our needs by using distance multiplied by area for the volume, so
W = P A(change in d)

Im wondering if I am going to have to use formulas related to specific heat capacities, but I am not quite sure. Any ideas?
(hint: when you see "monatomic" or "diatomic" mentioned in a problem there is likely going to be an issue involving heat capacity). You don't have to calculate the work done here.

This is an isobaric process since the force on the cylinder does not change. So how is T related to V?

If you remove 2093 J. how much does the temperature change? Work out final V from that.

AM
 
  • #3


I would like to clarify a few things before providing a response. Firstly, I assume that the gas is in a closed system and that the piston is the only external force acting on the gas. Secondly, I would like to clarify the units used in the problem. The area of the piston is given in m/s^2, which is not a standard unit for area. I will assume that it is meant to be m^2. Finally, I would like to mention that the equations mentioned in the post are correct and can be used to solve the problem.

Now, to answer the question, we can use the first law of thermodynamics, which states that the change in internal energy (U) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system. In this case, since the gas is in a closed system, the work done by the system is equal to the work done on the system by the piston, which can be calculated using the pressure and change in volume.

Therefore, we can write the equation as Uf - Ui = Q - P(Vf - Vi). Since the process is isobaric (constant pressure), we can simplify the equation to Uf - Ui = Q - P(AΔd), where A is the area of the piston and Δd is the change in distance.

We are given the heat removed from the gas, which is 2093 J, and the initial pressure of the gas, which is 1.01 x 10^5 Pa. We can also assume that the initial and final volumes are equal, as the piston is pushing down on the gas, causing it to compress and maintain the same volume. Therefore, we can write the equation as Uf - Ui = 2093 J - (1.01 x 10^5 Pa)(3.14 x 10^-2 m^2)(Δd).

To solve for the change in distance, we need to know the initial and final internal energies of the gas. This is where the specific heat capacity of the gas comes into play. We can use the equation U = nCvΔT, where n is the number of moles of gas, Cv is the specific heat capacity at constant volume, and ΔT is the change in temperature. Since the gas is monotomic, we can use the equation Cv = (3/2)R
 

1. What is the definition of heat removed from a gas in a cylinder?

Heat removed from a gas in a cylinder refers to the transfer of thermal energy from the gas to its surroundings, resulting in a decrease in the gas's temperature.

2. How is heat removed from a gas in a cylinder measured?

Heat removed from a gas in a cylinder is typically measured in joules (J) or calories (cal) using specialized equipment such as a calorimeter.

3. What factors affect the amount of heat removed from a gas in a cylinder?

The amount of heat removed from a gas in a cylinder depends on the initial temperature and pressure of the gas, the volume of the cylinder, and the type of gas present. External factors such as the ambient temperature and pressure can also impact the heat removal process.

4. How does the process of heat removal affect the gas in a cylinder?

As heat is removed from a gas in a cylinder, the gas molecules lose kinetic energy and move closer together, resulting in a decrease in volume and pressure. This can also lead to a decrease in the gas's temperature.

5. What are some real-world applications of heat removal from a gas in a cylinder?

The process of heat removal from a gas in a cylinder is used in various industries, such as refrigeration and air conditioning, to cool gases and liquids. It is also important in the study of thermodynamics and heat transfer in engineering and scientific research.

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