- #1

Karol

- 1,380

- 22

## Homework Statement

A cylindrical pump of length 25[cm] pumps air in atmoshpheric pressure ant temp 27

^{0}C into a big tank. the manometric pressure in the tank is 4[atm]. at which distance will the air start entering the tank.

## Homework Equations

In adiabatic process: ##P_1V_1^\gamma=P_2V_2^\gamma##

γ for air=1.4

## 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.