Question

A sample containing 321.5 g of nitrogen gas is expanded isothermally at T=298 K from 1.023dm^3 to 57.28dm^3. Calculate q, w, delta U and delta H for four processes assuming N^2 behaves ideally:

(a.) in a vacuum

(b.) against a constant external pressure equal to the final pressure of the sample

(c.) reversibly

(d.) adiabatically (non-isothermal)

Answer 1

a) In the case of expansion of any gas work is done by the system.

Now we know that

w = -p_extV = -p_ext (Vf - Vi)

where p external = external pressure ofthe system = 0

Thus the work done by the system will be:    w = 0( 57.28dm^3 - 1.023dm^3)

                                                                    = 0

Also q (heat) absorbed is the cosequence of gaining temperature.But gere it is at constant temperature.

q=-w

so q=0 hence no heat is absorbed.

As the process is isothermal, the change in internal energy=0

and deltaH= deltaU -p V

hence H = zero.

b) Given the amount of N2 is---------------321.5g = 321.5/28 moles

                                                                  = 11.5 moles

T= 298K

Vi = 1.023dm^3 = 1.023litres

We know the ideal gas equation: PV = nRT

Therefore Pi = nRT/Vi

                   = (11.5* 0.0821 lir.atm/mol K * 298)/1.023 lit

                   = 275 atm

We know that according to Boyles law:

P1V1=P2V2

therefore final pressure P2= P1V1/V2

                                      = (275 atm * 1.023 lit)/ 57.28lit

                                      = 4.91 atm    (THIS IS THE FINAL EXTERNAL PRESSURE)

Pext = 4.91 atm

Vf = 57.28lit

Vi=1.023 lit

Hence w= - Pext(Vf-Vi)

             = - 4.91atm(57.28-1.023)L

             = -276.2 atm L

In isothermal expansion: q=-w

                             therefore q= 276.2 atm.L

U = 0 and H = U + w

                         = -276.2 atm L

c) Reversible isothermal expansion : Then work done = -2.303 nRT log Vf/Vi

                                                     = - 2.303 * 11.5moles * 0.0821 L atm/mol K* 298K * log (57.28/1.023)

                                                     = - 1132.7 L atm

q= -w

   = 1132.7 L atm

delta U = 0 (isothermal)

delta H = delta + w

           = -1132.7 atm.L

d)Adiabatic process q= 0

Hence the change in internal energy = work done

We know that for an ideal gas:

T1V1^-1= T2 V2^-1

T2 = T1 (V1/V2)^-1 = 298 * (1.023/57.28)^1.4-1

                               = -208.4K

Hence work done= nR(T2-T1)/1-

                         = 11.5 * 0.0821 ( -208.4-298)/ (1-1.4)          since gamma value for a diatomic gas is 1.40

   = 1195.3 L atm

Delta U = 1195.3 L atm