Answer to Question #310543 in Molecular Physics | Thermodynamics for Vikram

Question

Consider a thermodynamics system consisting of 3 mol of an ideal gas occupying 0.03 m^3 volume at 300k temp. Determine its initial pressure. For this gas ,Y=1.4. It undegoes the following processes:

1.) It is compressed to 0.01 m^3 volume isothermally. Determine the pressure of the gas.

2.) Then it is allowed to expand adibatically, till it attains 1 atm pressure. Determine the final volume.(1atm=101325Nm^-2)

Draw a labelled indicator diagram of these processes.

Answer 1

Solution;

$n=3moles$

$V_1=0.03m^3$

$T_1=300K$

$\gamma=1.4$

From Ideal gas equation;

$PV=nRT$

Initial pressure is;

$P_1=\frac{nRT_1}{V_1}=\frac{3×8.314×300}{0.03}=249.42kPa$

1.After isothermal compression;

$P_1V_1=P_2V_2$

$P_2=\frac{P_1V_1}{V_2}=\frac{249.42×0.03}{0.01}=748.26kPa$

2.After adiabatic expansion;

$(\frac{P_2}{P_3})^{\frac1k}=\frac{V_3}{V_2}$

$V_3=0.01×(\frac{748.26}{101.325})^{\frac{1}{1.4}}$

$V_3=0.0417m^3$