Squeezing a piston and its effect on temperature

In summary, when a piston is compressed adiabatically, the work done on the gas will increase its internal energy, which causes an increase in temperature. This is due to the relationship between temperature, volume, and pressure in an ideal gas.
  • #1
qazxsw11111
95
0

Homework Statement


If a piston is compressed, how is the temperature affected?


Homework Equations


U=Q + W


The Attempt at a Solution



My soln: I thought that U=k.e. = (3/2)nRT=(3/2) pV ?
Since PV is constant (P1V1=P2V2), shouldn't temperature be constant?

Answer sheet: My answer sheet says that Q=0 since piston is insulated. and U=W. Since it is compressed, work is done on the gas, increasing its U, which increases T.

Which answer is correct?
 
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  • #2
Hi qazxsw11111,

qazxsw11111 said:

Homework Statement


If a piston is compressed, how is the temperature affected?


Homework Equations


U=Q + W


The Attempt at a Solution



My soln: I thought that U=k.e. = (3/2)nRT=(3/2) pV ?
Since PV is constant (P1V1=P2V2), shouldn't temperature be constant?

(P1V1=P2V2) is not always true; Boyle's law assumes that the temperature and amount of gas is kept constant.

If the amount of gas is constant, then for an ideal gas you would use:

[tex]
\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}
[/tex]

which follows directly from the ideal gas law.
 
  • #3


Both answers have elements of truth, but the answer sheet's explanation is more accurate. Let's break it down:

First, it is important to understand that temperature is a measure of the average kinetic energy of the molecules in a substance. When a piston is compressed, it reduces the volume of the gas inside, which means the molecules have less space to move around. This results in an increase in the average kinetic energy and therefore an increase in temperature.

In terms of the equations, the correct equation to use for this situation is the ideal gas law, which states that PV = nRT. This equation takes into account the relationship between pressure, volume, temperature, and the number of moles of gas. In this case, since the piston is being compressed (V decreases), and the number of moles of gas stays constant, the temperature must increase to maintain the equation's balance.

Your attempt at a solution is on the right track, but you are using the wrong equation. The equation U=Q + W is 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 plus the work (W) done on the system. This equation is useful for understanding the overall change in energy in a system, but it does not explain the relationship between compression and temperature.

The answer sheet's explanation is more accurate because it takes into account the fact that the piston is insulated, meaning that no heat (Q) is being added to the system. Therefore, the change in internal energy (U) is equal to the work (W) done on the system, which in this case is the compression of the piston. As mentioned before, this compression results in an increase in temperature.

In summary, when a piston is compressed, the temperature increases because the molecules in the gas have less space to move around, and therefore have higher average kinetic energy. The correct equation to use is the ideal gas law, and the answer sheet's explanation is more accurate because it takes into account the fact that the piston is insulated.
 

1. How does squeezing a piston affect the temperature of a gas?

When a piston is squeezed, the volume of the gas decreases, resulting in an increase in pressure. This increase in pressure causes the gas molecules to collide more frequently and with more force, increasing the kinetic energy and therefore the temperature of the gas. This is known as the ideal gas law.

2. Is there a limit to how much the temperature can increase when squeezing a piston?

According to the ideal gas law, as long as the volume and number of gas molecules are constant, the temperature will continue to increase with increasing pressure. However, at extremely high pressures, the gas may start to behave differently and the ideal gas law may no longer apply.

3. Can squeezing a piston cause a decrease in temperature?

Yes, in some cases squeezing a piston can cause a decrease in temperature. This is because when the gas is compressed, the molecules are forced closer together, causing them to interact more strongly and release heat. This is known as the Joule-Thomson effect.

4. Will the temperature increase be the same for all gases when squeezing a piston?

No, the temperature increase will depend on the specific gas being compressed. This is because different gases have different molecular properties, such as size and intermolecular forces, which can affect how they behave when compressed.

5. How does the initial temperature of the gas affect the temperature change when squeezing a piston?

The initial temperature of the gas does not have a direct effect on the temperature change when squeezing a piston. However, it can indirectly affect the temperature change by determining the initial kinetic energy of the gas molecules, which will impact how much the temperature increases when their kinetic energy increases due to compression.

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