Question

2) 2.5 mol of an ideal gas at 20 oC under 20 atm pressure, was expanded up to 5 atm pressure via; (a) adiabatic reversible and (b) adiabatic irreversible process. Calculate the values of w, q, ΔU, ΔH for each process. (Cv = 5 cal / mol.K ≈ 5/2 R; R ≈ 2 cal / mol.K) (Please find the desired values by making the corresponding derivations

Answer 1

Answer:

a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)

workdone (w) = -8442.6 J  ≈ -8.443 KJ

heat transferred (q) of the ideal gas = - w

q = 8.443 KJ

b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0

the workdone(w) in the ideal gas= - 4567.5 J  ≈ - 4.57 KJ

the heat transfer (q) of an ideal gas = 4.5675 KJ

Explanation:

given

mole of an ideal gas(n) = 2.5 mol

Temperature (T) = 20°C

= (20°C + 273) K  = 293 K

Initial pressure of the ideal gas(P₁) = 20 atm

Final pressure of the ideal gas(P₂) = 5 atm.

2) (a)for adiabatic reversible process,

note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.

Work done (w) = nRT ln$\frac{P_{1} }{P_{2} }$

= 2.5 mol × 8.314 J/mol K × 293 K × ln$\frac{5atm}{20atm}$

= 6090.01 J × [-1.3863]

= -8442.6 J  ≈ -8.443 KJ

So, the work done (w) of ideal gas = -8.443 KJ

For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0

From first law of thermodynamics:-

U = q + w

0 = q + w

q = - w

q = - (-8.443 KJ)

q = 8.443 KJ

heat transfer (q) of the ideal gas = 8.443 KJ

(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.

Work done (w) = -nRT(1 - ln$\frac{P_{1} }{P_{2} }$ )

= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]

= - 6090.01 J × 0.75

= - 4567.5 J  ≈ - 4.57 KJ

∴work done(w) of an ideal gas = - 4.57 KJ

For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0

From first law of thermodynamics:-

U = q + w

0 = q + w

q = - w

q = - (-4.5675 KJ)

q = 4.5675 KJ

the heat transfer (q) of an ideal gas = 4.5675 KJ