91影视

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

The length of cylinder$L_1$ = 0.250m

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

$C_v$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

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

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,

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

纬 is ratio of molar heat capacity

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

$C_V$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

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

Put all the values in equation (2)

Thus,

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

Therefore,

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

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 ${\mathrm{T}}_1=27.0\mathrm{C}掳$

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,

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

Put all the values in equation 3

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

Thus, the temperature of the compressed air is 196鈩�.

(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,

$W=Q鈭�\mathrm{螖}U$

In adiabatic process Q=0

$$\begin{gathered} W=0鈭�\mathrm{螖}U \\ W=鈭�\mathrm{螖}U \\ W=鈭�n{C}_V\mathrm{螖}T \\ W=鈭�n{C}_V\left(T_2鈭�T_1\right)鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�.\left(4\right) \end{gathered}$$

Put all the values in equation 4

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

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