Answer:
1/3p0
Explanation:
The combined gas law:
P1V1/T1 = P2V2/T2, where P, V and T are Pressure, Volume, and Temperature. Temperature must always be in Kelvin. The subscriopts 1 and 2 are for initial (1) and final (2) conditions.
In this case, temperature is constant (adiabatically). V1 = 2.0L and V2 = 6.0L. I'll assume P1 = p0.
Rearrange the combined gas law to solve for final pressure, P2:
P1V1/T1 = P2V2/T2
P2 = P1*(V1/V2)*(T2/T1) [Note how I've arranged the volume and temoperature terms - as ratios. This helps us understand what the impact of raising or lowering one on the variables will do to the system].
No enter the data:
P2 = P1*(V1/V2)*(T2/T1): [Since T2 = T1, the (T2/T1) terms cancels to 1.]
P2 = p0*(2.0L/6.0L)*(1)
P2 = (1/3)p0
The final pressure is 1/3 the initial pressure.
