Question

Nitrogen at an initial state of 300 K, 150 kPa and 0.2 m3 is compressed slowly in an isothermal process to a final pressure of 800 kPa. Determine the work done to the nitrogen and the amount of heat transfer during this process.

Answer 1

Work = - 4175.23 J

Heat Transfer = -4175.23 J

Step-by-step explanation:

The work done in an isothermal process, is given by the following formula:

W = RT ln (P1/P2)

where,

W = Work done

R = Universal Gas Constant = 8.314 J/mol.k

T = Constant Temperature = 300 K

P1 = initial pressure = 150 KPa

P2 = final pressure = 800 KPa

using these values, we get:

W = (8.314 J/mol.K)(300 k) ln (150/800)

W = - 4175.23 J

Here, negative sign shows that work is done on the system.

In isothermal process, from 1st law of thermodynamics:

Heat Transfer = Q = W (Since change in internal energy is zero for isothermal processes)

W = - 4175.23 J

Here, negative sign shows that heat is transferred from the system to surrounding.

Answer 2

Workdone = -50.22kJ

Heat transfer = -50.22kJ

Step-by-step explanation:

The following are given in the question;

Initial temperature, T₁ = 300K

Initial pressure, P₁ = 150kPa = 150 x 10³Pa

Initial volume, V₁ = 0.2m³

Final pressure, P₂ = 800kPa = 800 x 10³Pa

(a) Since the process is isothermal, there is no temperature change and the amount of work done (W) to the Nitrogen is given by;

W = P₁V₁ x ln (P₁ / P₂) -----------------(i)

Substitute the values of P₁, V₁ and P₂ into equation (i) as follows;

W = 150 x 10³ x 0.2 x ln [(150 x 10³) / (800 x 10³)]

W = 150 x 10³ x 0.2 x ln [(150) / (800)]

W = 150 x 10³ x 0.2 x ln [150 / 800]

W = 150 x 10³ x 0.2 x ln (0.1875)

W = 150 x 10³ x x 0.2 x (-1.674)

W = -50.22 x 10³

W = -50.22 kJ [-ve sign indicates that work is done on the system]

Therefore, the work done to the Nitrogen is -50.22kJ

(b) The internal energy, ΔU, is proportional to temperature. Now that the temperature is constant because the process is isothermal, this implies that the internal energy is zero(0).

Now using the first law of thermodynamics that states that;

ΔU = Q - W ----------------------(ii)

Where;

Q = Heat transfer

W = Work done

substitute ΔU = 0 into equation (ii) as follows;

0 = Q - Q

=> Q = W

Therefore, in an isothermal process, the amount of heat transfer (Q) in the process is equal to the work done (W) by or on the system

Therefore;

Q = -50.22kJ

The amount of heat transfer during the process is -50.22kJ