Ideal gases! For each of the following three scenarios, a-c, answer the followin

Question

Ideal gases! For each of the following three scenarios, a-c, answer the following three questions I-III. (This means there are NINE TOTAL ANSWERS here.)

Scenarios
a. Two moles of ideal gas are reversibly heated from 250 K to 300 K at a constant volume of 5L.
b. A mole of ideal gas is compressed reversibly from 100 L to 10 L at a constant temperature of 273 K.
c. One mole of ideal gas is reversibly adiabatically expanded from a temperature of 350 K and a volume of 1 L to a final volume of 5 L.

Questions
I. What is the change in entropy of the gas?
II. What is the change in entropy for the gas plus the heat reservoir which it contacts?
III. Is the process spontaneous? Briefly explain your answer.

Explanation / Answer

There are three types of processes in which entropy changes of an ideal gas. These three processes are:

1. Isothermal process
2. Isobaric process
3. Isochoric process

1. Isothermal process:
The process in which there is no change in temperature is known as Isothermal process. Entropy changes from S1 to S2 when gas absorbs heat during expansion. The heat taken by the gas is given by the area under the curve which represents the work done during expansion.

In other words, Q = W.
But Q = S2 S1 Tds = T(S2 – S1)
And W = P1 V1 ln (V2 / V1)
= R T1 ln (V2 / V1)

Therefore, T (S2 – S1) = R T1 ln (V2 / V1)
S2 – S1 = R ln (V2 / V1)
ST = R ln (V2 / V1)
= – R ln (P2 / P1)
= R ln (P1 / P2)

Hence, ST = R ln (V2 / V1)
= R ln (P1 / P2)
Hence, isothermal expansion of an ideal gas is accompanied by increase in entropy.

2. Isobaric Processes:
The process in which there is no change in pressure is known as Isobaric process. Consider an ideal gas at constant pressure and its temperature changes from T1 to T2 and entropy changes from S1 to S2.
Then, Q = Cp (T2 – T1)
Differentiating to find small increase in heat, dQ of the ideal gas when temperature rises is dT.

dQ = Cp.dT
dividing both sides by T, we will get:
dO / T = Cp. dT / T
dS = Cp. dT / T
Integrating both sides, we will get:
S2 S1 dS = Cp (T2 T1) dT / T
S2 – S1 = Cp ln (T2 / T1)
SP = Cp ln (T2 / T1)

Hence, increase in temperature of an ideal gas at constant pressure is accompanied by increase in entropy.

3. Isochoric Processes:
The process in which there is no change in volume is known as Isochoric process. Consider an ideal gas at constant volume and its temperature changes from T1 to T2 and entropy changes from S1 to S2.
Then, Q = Cv (T2 – T1)
Differentiating to find small increase in heat, dQ of the ideal gas when temperature rises is dT.
dQ = Cv .dT
dividing both sides by T, we will get:
dO / T = Cv. dT / T
dS = Cv. dT / T
Integrating both sides, we will get:
S2 S1 dS = Cv (T2 T1) dT / T
S2 – S1 = Cv ln (T2 / T1)
SP = Cv ln (T2 / T1)

Hence, increase in temperature of an ideal gas at constant volume is accompanied by increase in entropy of an ideal gas.

We can use Tds equation to solve for entropy change. The equation derived for simple compressible system, which undergo internal reversible process.
Tds= du + pdv
integrating from T1 to T2, with gas ideal assumption, constant heat capacity, and solve for ds

s2-s1 = Cv ln T2/T1 + R x ln v2/v1,

a. T1 = 250 K T2 = 300K and V1=V2=5L
= 0.833 + 8.314 . ln (5) = 1.0732 kJ/kmol.K

Entropy change per mole basis is 0.8732 kJ/kmol.K. The energy removed as heat can be find using dQ/ds = T.

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