A volume of ideal gas is first cooled without changing its volume and then expan

Question

A volume of ideal gas is first cooled without changing its volume and then expanded without changing its pressure, as shown by the path abc in the figure below.

1.Show that the temperature of the gas at a is the same as at c.

2.Compare the internal energy of the gas at a and c. Explain your answer.

3. I.How much heat does the gas exchange with its surroundings during the process abc?

II. Does the gas absorb heat or release heat during this process? Need the Explaination of your answer.

4.If the gas instead expands from state a to state c by the straight-line path as shown in the figure, how much heat does it exchange with its surroundings

Explanation / Answer

1. We know acc. to ideal gas law, PV = nRT

for a given fixed amount of gas at constant temp., we have the relation

P1V1 = P2V2... 1 and 2 represent diff. thermodynamic states of the gas.

At point a

PaVa = 6*10^5 * 0.02 = 12000 Nm or J

At point c

PcVc = 2*10^5 * 0.06 = 12000 Nm or J

As PaVa = PcVc, therefore temp. of the gas is same at a and c.

2. Since the internal energy of an ideal gas is a function of temperature only, the internal energy of the gas will be same at a and c because temp. is same at a and c as proved in part 1.

3. For a closed system undergoing changes in internal energy only, the first law of thermodynamics can be written as

dU = dQ + dW... U - internal energy, Q - heat exchange b/w system and surrounding, W - work done and d represents differetial change.

For process abc as dU = 0, because internal energy is same at a and c as explained above.

therefore dQ = -dW

Now work done can be calculated in 2 steps :-

Wab + Wbc

dW = -PdV

Wab = 0 because there is no volume change.

Wbc = P(V1 - V2)

= 2*10^5 (0.02 - 0.06) = -8000 J

therefore Q = 8000 J.

Acc. to the sign convention used here, + sign indicates that the heat is added to the system.

4. For the direct process ac

dU = dQ + dW

as already explained,dU = 0

dQ = -dW = PdV

here we will have to express P in terms of V. As it is a straight line it can be easily done.

P = -10^7V + 8*10^5

dQ =   (-10^7V + 8*10^5) dV

integrate this from 0.02 to 0.06.

Q = 16,000 J.

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