Question

2.1. Three moles of an ideal gas (with temperature-independent CP = (7/2)R, CV = (5/2)R) is contained in a horizontal piston/cylinder arrangement. The piston has an area of 0.1 m2 and mass of 500 g. The initial pressure in the piston is 101 kPa. Determine the heat that must be extracted to cool the gas from 375°C to 275°C at: (a) constant pressure; (b) constant volume.

Answer 1

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

$\frac{V_1}{V_2} = \frac{T_1}{T_2}$ 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