Question

With the pressure held constant at 250 kPa , 47 mol of a monatomic ideal gas expands from an initial volume of 0.70 m3 to a final volume of 1.9 m3 .

a) How much work was done by the gas during the expansion?

b) What were the initial temperature of the gas?

c) What were the final temperature of the gas?

d) What was the change in the internal energy of the gas?

e) How much heat was added to the gas?

Answer 1

given

Pa = 250KPa

v1 = 0.70m^3

v2 = 1.9m^3

n =47

we have the formula

Pa v1 = nRT1---------(1)

SO

1) T1 = Pav1 / nR

     =(250x10^3)(0.70) / (47)(8.31)

T1 = 448.06K (INITIAL TEMPERATURE)

SIMILARLY

2) T2 = Pav2 / nR

    =(250x10^3)(1.9) / (47)(8.310

T2 = 1216.1K (FINAL TEMPERATURE)

3) WORK DONE

dW = Pdv-----(2)

change in volume\

dv = v2 - v1

    =(1.9)-(0.70)

dv = 1.2m^3

pressure P = 1.013x10^5Pa

substituting in eqn (2) we have

dW = (1.013x10^5)(1.2)

dW = 121.56x10^3J

4)FROM THE FIRST LAW OF THERMODYNAMICS WE HAVE

CHANGE IN INTERNAL ENERGY IS

dU = dQ - dW------(3)

WE HAVE

dQ = 47(Cv)------(A)

here = T2 - T1

                = 1216.1 - 448.06

              = 768.04K

AND Cv= n /2 (R)

           =(47/2)(8.31)

Cv = 195.28J/mol/K

SUBSTITUTING THE VALUES OF Cv and in eqn (A)

we have

dQ = 47(195.28)(768.04)

dQ = 7.04x10^6J-----(B)

SUBSTITUTING THE VALUES OF dQ and dW in eqn(3) we have

dU = (7.04x10^6) - (121.56x10^3)

dU = 6.91x10^6J [ CHANGE IN INTERNAL ENERGY]

5) HEAT

Q = CvdT

     =(195.28)(768.04)

Q = 149.9x10^3J [AMOUNT OF HEAT ADDED]