Answer to Question #249466 in Mechanical Engineering for russman

Question

One kilogram of gas at an initial pressure of 0.11 MN/m² and a temperature of 15⁰C.

It is compressed isothermally until the volume becomes 0.1 m³ .Determine

a) the final pressure

b) the final temperature

c) the heat transfer

If the compression had been adiabatic, determine

d) the final pressure

e)the final temperature

f) the work transfer

Answer 1

Solution;

Given;

m=1kg

$P_1=0.11MN/m^2$

$T_1=15°c=288K$

$V_2=0.1m^2$

(a)

The final pressure;

From Ideal gas law;

$P_1V_1=mRT_1$

$V_1=\frac{mRT_1}{P_1}$ =$\frac{1×287×288}{0.11×10^6}$

$V_1=0.7514m^3$

For the two states;

$P_1V_1=P_2V_2$

$P_2=\frac{P_1V_1}{V_2}=\frac{0.11×10^6×0.7514}{0.1}$

$P_2=826540N/m^2$

(b)

Final temperature;

Since the process is isothermal ,the change in Temperature is 0,

Since;

$\Delta T=0$

$T_1=T_2$

$T_2=15°c=288K$

(c)

the heat transfer;

$Q=P_1V_1ln(\frac{V_2}{V_1})$

$Q=0.11×10^6×0.7514ln(\frac{0.1}{0.7514})$

$Q=-221844.4742J$

$Q=-221.844kJ$

If the process is adiabatic;

(d)

Final pressure;

Take n=1.4;

$PV^n=C$

$P_1V_1^n=P_2V_2^n$

$P_2=P_1(\frac{V_1}{V_2})^n$

$P_2=0.11×10^6(\frac{0.7514}{0.1})^{1.4}$

$P_2=1851877.914N/m^2$

$P_2=1.851MN/m^2$

(e)

Final temperature;

$\frac{T_1}{T_2}=(\frac{V_2}{V_1})^{n-1}$

$\frac{288}{T_2}=(\frac{0.1}{0.7514})^{0.4}$

$T_2=\frac{288}{0.4463}=645.27K$

$T_2=372.26°c$

(f)

Work transfer;

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{(1.851×0.1)-(0.11×0.7514)}{1.4-1}×10^6$

$W=256115J$

W=256.115kJ