Answer:
a) V1 = 5 ft^3
V2 = 2 ft^3
b) n = 1.378
Explanation:
Given data:
mass of gas = 4 lb
starting point
[tex]p_1 = 15 lbf/in^2[/tex]
[tex]v_1 = 1.25 ft^3/lb[/tex]
end point
[tex]p_2 = 60 lbf/in^2[/tex]
[tex]v_2 = 0.5 ft^3/lb[/tex]
Assume gas to be ideal
a) volume at point 1 [tex]= v_1 \times m[/tex]
[tex]V_1 = 1.25 \times 4 = 5 ft^3[/tex]
volume at point 2 [tex]= v_2 \times m[/tex]
[tex]V_2 = 0.5 \times 4 = 2 ft^3[/tex]
b) from ideal gas equation we have following equation
[tex]\frac{P_1}{P_2} = [\frac{V_2}{V_1}]^n[/tex]
taking log on both side of equation
[tex]ln [\frac{P_1}{P_2}] = n \times ln [\frac{V_2}{V_1}][/tex]
solving for n
[tex]n = \frac{ ln(\frac{15}{53})}{ ln(\frac{2}{5})}[/tex]
[tex]n = \frac{-1.262}{-0.916}[/tex]
n = 1.378
