Answer to Question #235097 in Molecular Physics | Thermodynamics for Unknown346307

Question

Air at 1.02 bar, 22°C, initially occupying a cylinder volume of 0.015 m3, is

compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate :

(i) The final temperature ;

(ii) The final volume ;

(iii) The work done.

Answer 1

Solution;

Give;

$p_1=1.02 bar$

$T_1=22°c=295K$

$V_1=0.015m^3$

$p_2=6.8bar$

Law of compression;

$pV^\gamma=C$

(i)Final temperature;

Using the relation;

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

$\gamma$ for air is 1.4;

$\frac{T_2}{295}=(\frac{6.8}{1.02})^{\frac{1.4-1}{1.4}}$

$T_2=295(\frac{6.9}{1.02})^{\frac{0.4}{1.4}}$ =507.24K

Final temperature is;

507.24-273=234.24°c

(ii)Final volume;

Using the relation;

$p_1V_1^\gamma=p_2V_2^\gamma$

Resolve as;

$\frac{V_2}{V_1}=(\frac{p_1}{p_2})^\frac{1}{\gamma}$

$V_2=0.015×(\frac{1.02}{6.8})^\frac{1}{1.4}$

$V_2=0.00387m^3$

(iii)Work done;

Using;

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

m is the mass found by the following relation;

$pV=mRT$

$m=\frac{p_1V_1}{RT_1}$ =$\frac{1.02×10^5×0.015}{287×295}$

$m=0.01807kg$

Hence;

$W=\frac{0.01807×287×(295-507.24)}{1.4-1}$

$W=-2751J$

Negative sign means work is done on the air.