# Squeezing a piston and its effect on temperature

1. Sep 13, 2008

### qazxsw11111

1. The problem statement, all variables and given/known data
If a piston is compressed, how is the temperature affected?

2. Relevant equations
U=Q + W

3. 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), shouldnt 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.

2. Sep 13, 2008

### alphysicist

Hi qazxsw11111,

(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.