Answer to Question #233158 in Molecular Physics | Thermodynamics for Unknown346307

Question

20 g of gas at 20oC and 1 bar pressure is compressed to 9 bar by the law pV1.4 = C.

Taking the gas constant R = 287 J/kg K calculate the work done. (Note that for a compression

process the work will turn out to be positive if you correctly identify the initial and final

conditions)

Answer 1

Given;

$m=20g=0.02kg$

$T_1=20°c=293K$

$p_1=1 bar$

$p_2=9 bar$

By the law;

$pV^{1.4}=C$

We have;

$p_1V_1^{1.4}=p_2V_2^{1.4}$

$T_2=T_1×(\frac{p_2}{p_1})^{1-\frac{1}{1.4}}$

$T_2=293×9^{0.286}$ =549K

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

$W=\frac{0.02×287×(549-293)}{1.4-1}$

$W=3.68kJ$