Answer to Question #233817 in Molecular Physics | Thermodynamics for Unknown346307

Question

7. An air compressor commences compression when the cylinder contains 12 g at a pressure is 1.01

bar and the temperature is 20o

C. The compression is completed when the pressure is 7 bar and

the temperature 90o

C.

The characteristic gas constant R is 287 J/kg K. Assuming the process is reversible and

polytropic, calculate the index of compression and the work done.

Answer 1

Solution;

Given;

$m=12g=0.012kg$

$p_2=1.01$ bar

$T_1=20°c=293K$

$p_2=7$ bar

$T_2=90°c=363K$

$R=287J/kgK$

To find index of compression;

Since;

$p_1V_1^n=p_2V_2^n$

In terms of T;

$\frac{T_2}{T_1}=(\frac{p_2}{p_1})^{1-\frac1n}$

Hence;

$\frac{363}{293}=(\frac{7}{1.01})^{1-\frac1n}$

$1.2389=6.9307^{1-\frac1n}$

$1-\frac1n=\frac{ln(1.2389)}{ln(6.9307)}$ =$0.1107$

$n=1.124$

To find the work done;

$W=\frac{mR(T_2-T_1)}{n-1}$

$W=\frac{0.012×287×(363-293)}{1.124-1}$

$W=1944.2J$