Thermodynamic Processes

Two moles of a monatomic ideal gas at (5 MPa, 5 L) is expanded isothermally until the volume is doubled (step 1). Then it is cooled isochorically until the pressure is 1 MPa (step 2). The temperature drops in this process. The gas is now compressed isothermally until its volume is back to 5 L, but its pressure is now 2 MPa (step 3). Finally, the gas is heated isochorically to return to the initial state (step 4). (a) Draw the four processes in the pV plane. (b) Find the total work done by the gas.

Respuesta :

Answer:

A. Part a is the attachment

B. total work = 10.4kj

Explanation:

[tex]workdone=nRT1ln\frac{Vb}{Va}[/tex]

T1 = constant temperature

nRT1 = PaVa = PbVb

We write equation as

[tex]workdone =(PaVa)ln\frac{Vb}{Va}[/tex]

5ma = Pa, 5L = Va, Vb = 10L(temperature is doubled)

[tex]w1 = workdone =(5mpa*5L)ln\frac{10L}{5L}[/tex]

W1 = 25 ln2

W1 = 25 x 0.693

= 17.327kj

The isochoric expansion has no change in volume. So,

W2 = 0

Isothermal compression

[tex]w3=nRT3ln\frac{Vd}{Vc}[/tex]

T3 = constant temperature

nRT3 = PcVc = PdVd

[tex]workdone=(PcVc)ln\frac{Vd}{Vc}[/tex]

Pc = 1mpa Vc = 10L Vd = 5L

[tex]w3=(1)(10)ln\frac{5L}{10L}[/tex]

= 10x-0.693

= -6.93kj

Isochoric compression has no change in volume. Workdone w4 = 0

Total workdone = w1 + w2 + w3 + w4

= 17.33 + 0 + (-6.93) + 0

= 10.4kj

Ver imagen ogorwyne
ACCESS MORE
ACCESS MORE
ACCESS MORE
ACCESS MORE