Question:
Calculate the energy required to change the temperature of 1.00 kg of ethane (C2H6) from 25.0"C to 73.4"C in a rigid vessel. (Cv for C2H6 is 44.60 J K%1 mol%1.)
Calculate the energy required for this same temperature change at constant pressure.
Calculate the change in internal energy of the gas in each of these processes.
Answer:
The answers are
qv =71787.16 J
qp = 85169.99 J
ΔE = 71787 J for both constant volume and constant pressure processes
Explanation:
Mass of ethane gas, m = 1.00 kg = 1000 g
Cv = 44.60 J/(K·mol)
The temperature change is given by
ΔT T₂ - T₁ = 73.4 °C - 25 °C = 48.4 °C
The molar mass of ethane = 30.07 g/mol
Number of moles, n of ethane = mass/(molar mass) = 1000/30.07
= 33.26 moles
Therefore the energy required at constant volume to change the temperature of 1.00 kg of ethane (C₂H₆) from 25.0 °C to 73.4 °C
is given by H qv= n ·Cv·ΔT = 33.26× 44.60×48.4 = 71787.16 J
b) At constant pressure we have Cp - Cv = R
∴ Cp = R + Cv
= 8.3145 + 44.6 = 52.9145 J/(K·mol)
Therefore we have H, qp = n ·Cp·ΔT = 33.26×52.9145×48.4
= 85169.99 J
For constant volume process, we have
q = U
Therefore ΔU = Δq = 71787 J
For constant pressure process, we have
ΔU = n×Cv×dT = 71787 J
