Answer to Question #273792 in Molecular Physics | Thermodynamics for EFIE

Question

While the temperature remains constant at 38 deg. C, 2.5 kg of a certain ideal gas whose Cp = 1.1724 kJ/kg-K, Cv = 0.8375 kJ/kg-K rejects 300kJ of heat. If p1 = 586kPa, determine:

(a) the final pressure in kPa

(b) the initial volume in cubic meter

(c) the final volume in cubic meter

Answer 1

Solution;

Given;

$T_1=T_2=38°c=311K$

m=2.5

$Q=300kJ$

$P_1=586kPa$

$C_p=1.1724kJ/kgK$

$C_v=0.8375kJ/kgK$

$R=C_p-C_v=0.3349kJ/kgK$

From;

$PV=mRT$

We have;

$V_1=\frac{2.5×0.3349×311}{586}=0.4443m^3$

$Q=p_1v_1ln(\frac{v_2}{v_1})$

$300=586×0.4443×ln(\frac{v_2}{0.4443})$

$V_2=0.1404m^3$

(a)

Final pressure in kPa

$P_2=\frac{V_1}{V_2}P_1=\frac{0.4443}{0.1404}×586=1854.4kPa$

(b)

Initial volume in m³;

$V_1=0.4443m^3$

(c)

Final volume;

$V_2=0.1404m^3$