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

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.

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

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml