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3 years ago

Thermodynamic Processes

Physics

1 answer:

goblinko [34]3 years ago

5 0

Answer:

A. Part a is the attachment

B. total work = 10.4kj

Explanation:

$workdone=nRT1ln\frac{Vb}{Va}$

T1 = constant temperature

nRT1 = PaVa = PbVb

We write equation as

$workdone =(PaVa)ln\frac{Vb}{Va}$

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

$w1 = workdone =(5mpa*5L)ln\frac{10L}{5L}$

W1 = 25 ln2

W1 = 25 x 0.693

= 17.327kj

The isochoric expansion has no change in volume. So,

W2 = 0

Isothermal compression

$w3=nRT3ln\frac{Vd}{Vc}$

T3 = constant temperature

nRT3 = PcVc = PdVd

$workdone=(PcVc)ln\frac{Vd}{Vc}$

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

$w3=(1)(10)ln\frac{5L}{10L}$

= 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

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Answer:

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