Squeezing a piston and its effect on temperature

AI Thread Summary
Compressing a piston increases the temperature of the gas inside due to the work done on it, as indicated by the first law of thermodynamics (U=Q+W). When the piston is insulated, heat transfer (Q) is zero, meaning all work done (W) contributes to the internal energy (U) of the gas, thus raising its temperature. The assumption that pressure and volume remain constant is incorrect in this scenario; instead, the ideal gas law must be applied to relate pressure, volume, and temperature changes. Therefore, as the piston compresses the gas, the temperature will rise, contrary to the initial assumption that it would remain constant. Understanding these principles is crucial for accurately analyzing thermodynamic processes involving gases.
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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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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:

<br /> \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}<br />

which follows directly from the ideal gas law.
 
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