# Pump compresses air into a tank

1. Jan 4, 2015

### Karol

1. The problem statement, all variables and given/known data
A cylindrical pump of length 25[cm] pumps air in atmoshpheric pressure ant temp 270C into a big tank. the manometric pressure in the tank is 4[atm]. at which distance will the air start entering the tank.

2. Relevant equations
In adiabatic process: $P_1V_1^\gamma=P_2V_2^\gamma$
γ for air=1.4

3. The attempt at a solution
I assume an adiabatic process. i think that manometric pressure is gauge pressure:
$$P_1V_1^\gamma=P_2V_2^\gamma\rightarrow V_2=\sqrt[\gamma]{\frac{P_1}{P_2}V^{\gamma}}$$
$$\rightarrow \ln V_2=\ln\left(\frac{P_1}{P_2}V_1^{\gamma}\right)^{\frac{1}{\gamma}}=\frac{1}{\gamma}\ln\left(\frac{P_1}{P_2}V_1^{\gamma}\right)$$
$$\ln V_2=\frac{1}{\gamma}\left[\ln\left(P_1V_1^{\gamma}\right)-\ln P_2\right]=\frac{1}{\gamma}\left[\ln P_1+\gamma\ln V_1-\ln P_2\right]$$
The volume is area A times length x. for our data:
$$\ln V_2=\ln A+\ln x=\frac{1}{1.4}\left[\ln 0+1.4\ln (A\cdot 25)-\ln 5\right]=\frac{1}{1.4}\left[1.4(\ln A+\ln 25)-\ln 5\right]$$
$$\ln A+\ln x=\ln A+\ln 25-\ln 5\rightarrow \ln x=\ln\left(\frac{25}{5}\right)=\ln 5\rightarrow x=5[cm]$$
It should be 7.9[cm] before the end of the cylinder.
If i don't assume adiabatic process and only use $PV=nRT$ i have 2 unknowns: the distance and the final temperature.

2. Jan 4, 2015

### Staff: Mentor

You forgot to divide ln(P2) by 1.4.
Also, P1 = 1, so you shouldn't have ln(0) in the equation.
Also, it was not necessary to work this problem in terms of natural logs.

Chet

3. Jan 4, 2015

### Karol

$$\ln A+\ln x=\ln A+\ln 25-\frac{\ln 5}{1.4}\rightarrow x=7.9[cm]$$
Which other way besides natural logs?

4. Jan 4, 2015

### Staff: Mentor

$$\frac{V_2}{V_1}=\left(\frac{P_1}{P_2}\right)^{1/\gamma}$$
$$\frac{x}{25}=0.2^{(1/\gamma)}$$
Chet

5. Jan 5, 2015

Thanks Chet