Question

The initial state of a quantity of monatomic ideal gas 2 atm, 10 liter, and 373K. The gas is isothermally expanded to a volume of 20 liters and is then cooled at constant pressure to the volume of V3p. This volume is such that a reversible adiabatic compression to a pressure of 1 atm returns the system to its initial state. (1) the V3p of the system. (2) the work done by or on the system. (3) the total heat entering or leaving the system. (4) the change in the internal energy and (5) Enthalpy (6) change in the Entropy.

Answer 1

First state: P1 = 2atm. T1 = 373K, V1 = 10litres

1-2: Isothermal Expansion.

Second state: P2= P2, T2 = 373K, V2 = 20litres

2-3: Isobaric Cooling.

Third state: P3 = P2, T3 = T3, V3 = V3

3-1: Adiabatic Compression.

Now, since 1-2 is an isothermal process, we can use the following relation:

Therefore, P2 = 1 atm

So, P3 = P2 = 1 atm

Now, 3-1 is an adiabatic process, So we can use the following relation

Since, it is given that it is a monoatomic gas,

We know that P1 = 2atm, P3 = 1 atm, V1 = 10 litres, T1 = 373K.

Substituting the values in the above equation, we get:

V3 = 15.15 litres, T3 = 282.68 K.

Now, we will calculate the work done in different processes.

For process 1-2: W =

Substituting values, we get:

W1 = 13.863 Joules

For process 2-3, W = P2*(V3-V2)

Substituting values, we get:

W2 = -4.85 Joules

For process 3-1, W=

Substituting values, we get:

W3 = -7.275 Joules

So, Net Work during the process = 13.386-4.85-7.275 = 1.738 Joules.

So, the work done by the system is 1.738 Joules.

Since the combination of all processes is a cyclic process,

Change in internal energy is 0.

Now, according to the first law of thermodynamics,

Therefore, Q= 0+1.738 = 1.738 Joules.

So, heat entering the system is 1.738 Joules.