How Does Gearbox Volume Change at 353 K?

[M1342]

What is the volume in state 2
The volume of the gearbox is 0,00026 m3 with lubrication and air. The volume of the lubrication is 0.0001139 m3 and expands by 13,1% in state 2 which is 353 K, the rest is air.
I traid to calculate P2 and get the following.
By equation [V1*P1/T1] = [V2*P2/T2]
V1= 0.000260 m3
P1=1 bar
T1=293 K
T2=353
V2=V1- (Air+lubrication-V1), in state 2.
Air: V1/T1 = V2/T2
V1 [air] = 0.000260 m3 – 0.0001139 m3 = 0.000146 m3
It gives V2 [air]= 0.000146*353/293=0.000176 m3
Lubrication: 0.0001139*1.131=0.000129 m3
V2=0.000260 – ((0.000146 + 0.000129)-0.000260) = 0.000245 m3
It gives P2 = (V1*P1*T2) / (T1*V2) = 1.278 bar
Is my calculations wrong, I’m very uncertain about V2.

 

[Valkarie]

Your calculations are correct. The volume of the gearbox in State 2 is 0.000245 m3 and the pressure is 1.278 bar.