Answer:
The answers to the question are
The heat that must be extracted to cool the gas from 375°C to 275°C at
(a) Constant pressure Q = -6110.78 J
(b) Constant volume Q = -4365 J
Explanation:
Cp = (7/2)R
CV = (5/2)R)
Area of piston = 0.1 m²
Mass of piston, m = 500 g =0.5 kg
Mass of piston = 500 g
Initial pressure of piston p₁ = 101 kPa
Initial temperature of piston = 375 °C = 648.15 K
Final temperature of piston =275 °C = 548.15 K
Pressure from the piston = m×g/0.1 =0.5×9.81/0.1 = 49.05 Pa
Pressure from gas alone = 101000-49.05= 100950.95 Pa
Therefore we have from the general gas equation
P₁V₁ = n·R·T₁ Where
V₁ = Initial volume piston = n·R·T₁/P₁ R = 8.314 J / mol·K.
V₁ = (8.314×2.1×648.15)/100950.95 = 0.112097 m³
For V₂ =n·R·T₂/P₂ = (8.314×2.1×548.15)/101000 = 0.094756 m³
V₂ = T₂/T₁×V₁ = 548.15/648.15×0.11204 = 0.09476 m³
But p·V = m·R·T ⇒ m·R = P·V/T =100950.95*0.112097/648.15 = 17.45938
For constant pressure we have Q = m*cp*(T₂-T₁) =7/2·R*m*(T₂-T₁) =
7/2*17.45938*(548.15 -648.15) = -6110.7838 J =
(b) At constant volume, the heat that must be extracted =
We have P₁/T₁ = P₂/T₂
m*cv*(T₂-T₁)
= 5/2·R*m*(548.15-648.15) =5/2*17.46*(-100) = -4365 J