Respuesta :
Answer:
Explanation:
Given
[tex]P_1=150 kPa[/tex]
[tex]T_1=20^{\circ}C[/tex]
[tex]V_1=0.5 m^3[/tex]
[tex]T_2=140^{\circ}C[/tex]
[tex]P_2=400 kPa[/tex]
R for Helium [tex]R=2.076[/tex]
[tex]c_v=3.115 kJ/kg-K[/tex]
mass of gas [tex]m=\frac{P_1V_1}{RT_1}[/tex]
[tex]m=\frac{150\times 0.5}{2.076\times 293}[/tex]
[tex]m=0.123 kg[/tex]
Similarly [tex]V_2[/tex] can be found
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
[tex]V_2=0.264 m^3[/tex]
Work done [tex]W=\int_{V_1}^{V_2}PdV[/tex]
[tex]W=\frac{P_2V_2-P_1V_1}{n-1}[/tex]
[tex]W=\frac{mR(T_2_T_1)}{n-1}[/tex]
Since it is a polytropic Process
therefore [tex]PV^n=c[/tex]
[tex]P_1V_1^n=P_2V_2^n[/tex]
[tex](\frac{V_1}{V_2})^n=\frac{P_2}{P_1}[/tex]
[tex](\frac{0.5}{0.264})^n=\frac{400}{150}[/tex]
[tex]n=\frac{\ln 2.66}{\ln 1.893}[/tex]
[tex]n=1.533[/tex]
[tex]W=\frac{0.123\times 2.076(140-20)}{1.533-1}[/tex]
[tex]W=57.48 kJ[/tex]
From Energy balance
[tex]E_{in}-E_{out}=\Delta E_{system}[/tex]
Neglecting kinetic and Potential Energy change
[tex]Q_{in}+W_{in}=change\ in\ Internal\ Energy[/tex]
Change in Internal Energy [tex]\Delta U=u_2-u_1[/tex]
[tex]\Delta U=mc_v(T_2-T_1)[/tex]
[tex]\Delta U=0.123\times 3.115(140-20)[/tex]
[tex]\Delta U=45.977 kJ[/tex]
[tex]Q_{in}+57.48=45.977[/tex]
[tex]Q_{in}=-11.50 kJ[/tex]
i.e. Heat is being removed
