An air pump has a cylinder 0.250 m long with a movable piston. The pump is used to compress air from the atmosphere (at absolute pressure 1.01 * 105 Pa) into a very large tank at 3.80 * 105 Pa gauge pressure. (For air, CV = 20.8 J>mol # K.) (a) The piston begins the compression stroke at the open end of the cylinder. How far down the length of the cylinder has the piston moved when air first begins to flow from the cylinder into the tank? Assume that the compression is adiabatic. (b) If the air is taken into the pump at 27.0°C, what is the temperature of the compressed air? (c) How much work does the pump do in putting 20.0 mol of air into the tank?

Cylinder with initial pressure equals to atmospheric pressure$1.01\times {10}^5\mathrm{Pa}$

The length of cylinder$L_1$ = 0.250m

A piston compresses air to fill tank with pressure =$3.8\times {10}^5\mathrm{Pa}$

$C_v$of the air = 20.8 kJ/mol.

(a)the length of the cylinder that the piston has moved.

The final pressure $p_2$act on the air will be,

$p_2$= pressure of tank + pressure of air

$=3.8\times {10}^5\mathrm{Pa}+1.01\times {10}^5\mathrm{Pa}=4.81\times {10}^5\mathrm{Pa}$

Calculate final length of the air ${\mathrm{L}}_2$then calculate moved distance by ${\mathrm{L}}_1-{\mathrm{L}}_2$

The final and initial state of pressure and volume is given as,

${\mathrm{p}}_1{\left({\mathrm{v}}_1\right)}^{\mathrm{y}}={\mathrm{p}}_2{\left({\mathrm{v}}_2\right)}^{\mathrm{y}}$……………………………………………………………………. (1)

γ is ratio of molar heat capacity

$\gamma =\frac{C_P}{C_V}=\frac{C_V+R}{C_V}=1+\frac{R}{C_V}=1.40$

$C_V$is given as 20.8J/mol

R is gas constant =8.314J/mol

Volume of cylinder=area of base x length

So,

$$\begin{gathered} {\mathrm{p}}_1={\left({\mathrm{AL}}_1\right)}^{\mathrm{y}}={\mathrm{p}}_2{\left({\mathrm{AL}}_2\right)}^{\mathrm{y}} \\ {\mathrm{p}}_1{\left({\mathrm{L}}_1\right)}^{\mathrm{y}}={\mathrm{p}}_2{\left({\mathrm{AL}}_2\right)}^{\mathrm{y}} \\ {\left(\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}\right)}^{\mathrm{y}}=\frac{{\mathrm{p}}_1}{{\mathrm{p}}_2} \end{gathered}$$

Solve for

${\mathrm{L}}_2={\mathrm{L}}_1{\left(\frac{{\mathrm{p}}_1}{{\mathrm{p}}_2}\right)}^{\frac{1}{\mathrm{y}}}$ …………………………………………………………………………….(2)

Put all the values in equation (2)

Thus,

$$\begin{gathered} {\mathrm{L}}_2=0.250\mathrm{m}{\left(\frac{1.01\times {10}^5\mathrm{Pa}}{4.81\times {10}^5\mathrm{Pa}}\right)}^{\frac{1}{1.40}} \\ {\mathrm{L}}_2=0.082\mathrm{m} \end{gathered}$$

Therefore,

${\mathrm{L}}_1‐{\mathrm{L}}_2=0.250\mathrm{m}-0.082\mathrm{m}=0.168\mathrm{m}$

Thus, the length of the cylinder that the piston moves when air first begins to flow from the cylinder into the tank is 0.168m.

(b)the temperature of the compressed air

Initial temperature ${\mathrm{T}}_1=27.0\mathrm{C}°$

To find final temperature ${\mathrm{T}}_2$When it is compressed the final and initial state of temperature and volume are used for adiabatic processes.

So,

$$\begin{gathered} T_1{\left(V_1\right)}^{\gamma -1}=T_2{\left(V_2\right)}^{\gamma -1} \\ {T}_1{\left(A_1\right)}^{\gamma -1}=T_2{\left(A{L}_2\right)}^{\gamma -1} \\ {T}_1{\left(L_1\right)}^{\gamma -1}=T_2{\left(L_2\right)}^{\gamma -1} \\ {T}_2=T_1{\left(\frac{L_1}{L_2}\right)}^{\gamma -1}.................................................................\left(3\right) \end{gathered}$$

Put all the values in equation 3

$$\begin{gathered} T_2=300{\left(\frac{0.250m}{0.082m}\right)}^{1.40-1} \\ {T}_2=468\mathrm{K}=196C^0 \end{gathered}$$

Thus, the temperature of the compressed air is 196℃.

(c)The work pump does by putting 20.0 mol of air into the tank.

Take number of moles of air n=20.0mol and according to first law of thermodynamics,

$W=Q-\mathrm{\Delta }U$

In adiabatic process Q=0

$$\begin{gathered} W=0-\mathrm{\Delta }U \\ W=-\mathrm{\Delta }U \\ W=-n{C}_V\mathrm{\Delta }T \\ W=-n{C}_V\left(T_2-T_1\right)\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(4\right) \end{gathered}$$

Put all the values in equation 4

$$\begin{gathered} W=-n{C}_V\left(T_2-T_1\right) \\ W=-20.0\times 20.8(468-300)=-70.1\mathrm{kJ} \end{gathered}$$

Thus, the work pump does in putting 20.0 mol of air into the tank is -70.1kJ.