Squeezing a piston and its effect on temperature

[qazxsw11111]

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?

 

[alphysicist]

Hi qazxsw11111,

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:

$$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$$

which follows directly from the ideal gas law.