# Ideal gas pressure reduction

1. Jul 21, 2007

### trah22

1. The problem statement, all variables and given/known data
an ideal gas is allowed to expand adiabatically until its volume increases by 50%. By what factor is the pressured reduced?(gamma=5/3)

2. Relevant equations

P1V1^gamma=P2V2^gamma

3. The attempt at a solution

P1V1^gamma=P2V2^gamma

v2=v1+.5V1=3/2V1

(p1/P2)=(v2/v1)^gamma=(3/2v1/v1)^gamma=3/2^gamma

so P2=P1/(3/2)^gamma=0.51P1 sp P2=.5P1

so reduced by approximately factor of 2

This is actually the correct answer, but i dont understand the why the intial setup of the calculation, is P1/P2=(V2/V1)^gamma, the problem is looking for the reduction of the pressure, could someone explain why your suppose to set up calculation this way?

2. Jul 21, 2007

### trah22

more specifically why do you put p1 over p2 (p1/p2) equal to v2 over v1 (v2/v1) in solving this problem?

3. Jul 21, 2007

### Andrew Mason

?? It is just algebra:

$$P_1V_1^\gamma = K = P_2V_2^\gamma$$

$$\frac{P_1V_1^\gamma}{P_2} = V_2^\gamma$$

$$\frac{P_1}{P_2} = \frac{V_2^\gamma}{V_1^\gamma}= \left(\frac{V_2}{V_1}\right)^\gamma$$

AM

4. Jul 21, 2007

### trah22

lol for some reason i thought something else was going on, thanks anyways