An ideal gas undergoes a reversible isothermal expansion at 77 . 0 ° C , increasing its volume from 1 . 30 L to 3 . 90 L . The entropy change of the gas is 22 . 0 J K - 1 . How many moles of gas are present?

Given: Process undergoes an isothermal expansion at temperature, T = 77 ° C = 350 K Initial volume, v i = 1 .30 L Final volume, v f = 3 .90 L Change in entropy, Δ S = 22 .0 J K - 1 Let us assume the number of mole to be n , Since the process is isothermal, change in entropy is, Δ S = n R ln v f v i (Where R = Universal gas constant = 8 . 314 J K - 1 mol - 1 ) Substituting all the values in the above formula, ⇒ 22 = n × 8 .314 × ln 3 .90 1 .30 ⇒ n = 2 .2 8 .314 × ln 3 = 0 . 2408 mol

Suppose 1 . 00 mol of a monatomic ideal gas is taken from initial pressure p 1 and volume V 1 , through two steps: ( 1 ) an isothermal expansion to volume 2 . 00 V 1 and ( 2 ) a pressure increase to 2 . 00 p 1 at constant volume. What is Q p 1 V 1 for ( a ) step 1 and ( b ) step 2 ? What is W p 1 V 1 for ( c ) step 1 and ( d ) step 2 ? For the full process, what are ( e ) Δ E int p 1 V 1 and ( f ) Δ S ? The gas is returned to its initial state and again taken to the same final state but now through these two steps: ( 1 ) an isothermal compression to pressure 2 . 00 p 1 and ( 2 ) a volume increase to 2 . 00 V 1 at constant pressure. What is Q p 1 V 1 for ( g ) step 1 and ( h ) step 2 ? What is W p 1 V 1 for ( i ) step 1 and ( j ) step 2 ? For the full process, what are ( k ) Δ E int p 1 V 1 and ( l ) Δ S ?

Given: Number of moles of monoatomic gas, n = 1 .00 mol For monoatomic gas, C v = 3 2 R , C p = 5 2 R The given process can be shown on the following P - V diagram, ( a ) For step 1 : Heat transfer is equal to the work done, i.e., Q = W for isothermal process 1 → 2 and work done for isothermal process is, W = P 1 V 1 ln V 2 V 1 Also, heat transfer is, Q = P 1 V 1 ln V 2 V 1 ⇒ Q P 1 V 1 = ln 2 V 1 V 1 = ln 2 = 0 .693 For step 2 , the process is isochoric. So, heat transfer is, Q = n C v d T = n × 3 2 R T 3 − T 2 ⇒ Q = n × 3 2 × R × T 2 T 3 T 2 − 1 ⇒ Q = 3 2 × n R T 1 T 3 T 2 − 1 ∵ T 2 = T 1 ⇒ Q = 3 2 × P 1 V 1 T 3 T 2 − 1 ∵ P 1 V 1 = n R T ⇒ Q = 3 2 × P 1 V 1 P 3 P 2 − 1 . . . 1 ( b ) For process 1 → 2 , from Boyle's law, P 1 V 1 = P 2 V 2 ⇒ P 1 V 1 = P 2 × 2 V 1 ⇒ P 2 = P 1 2 And P 3 = 2 P 1 ∴ P 3 P 2 = 2 P 1 P 1 / 2 = 4 . . . 2 ⇒ T 3 T 2 = 4 On putting equation 2 in equation 1 , we have, ⇒ Q = 3 2 P 1 V 1 4 − 1 ⇒ Q P 1 V 1 = 3 2 × 3 = 9 2 = 4 .50 ( c ) For step ( 1 ) : The process 1 → 2 is isothermal, so work done is, W = P 1 V 1 ln V 2 V 1 ⇒ W P 1 V 1 = ln 2 V 1 V 1 = ln 2 = 0 .693 ( d ) For step 2 , 2 → 3 : The process is isochoric, so work done, W = 0 and W P 1 V 1 = 0 ( e ) For full process, change in internal energy is, Δ E int = Δ E 1 → 2 + Δ E 2 → 3 . . . 4 Process 1 → 2 is isothermal, so Δ E 1 → 2 = 0 Process 2 → 3 is isochoric, so Δ E 2 → 3 = n C v T 3 − T 2 and is same as in equation 3 , so ⇒ Q = 3 2 P 1 V 1 4 − 1 = Δ E 2 → 3 . . . 5 On putting equation 5 in equation 4 , we get, ⇒ Δ E int = 0 + 3 2 P 1 V 1 × 3 ⇒ Δ E int P 1 V 1 = 3 2 × 3 = 9 2 = 4 .5 ( f ) Change in entropy is Δ S for the total process. Δ S = Δ S 1 → 2 + Δ S 2 → 3 And Δ S 1 → 2 = n R ln V 2 V 1 and Δ S 2 → 3 = n C v ln T 3 T 2 ⇒ Δ S 1 → 2 = n R ln 2 V 1 V 1 a n d Δ S 2 → 3 = n C v ln 4 Substituting all the values for Δ S , we get, = Δ S = n R ln 2 + n C v ln 4 ⇒ Δ S = 1 × 8 .312 × ln 2 + 1 × 3 2 × 8 .314 × ln 4 ⇒ Δ S = 5 .763 + 17 .288 = 23 .051 J K - 1 In another case, the gas is returned to initial state and then moved through an isothermal compression and isobaric process, as shown in the following P - V diagram, ( g ) For Q P 1 V 1 : For isothermal process, heat transfer, Q = P 1 V 1 ln V 2 V 1 ⇒ Q = P 1 V 1 ln P 1 P 2 ∵ P 1 V 1 = P 2 V 2 ⇒ Q = P 1 V 1 ln 1 2 = − P 1 V 1 ln 2 ⇒ Q P 1 V 1 = − ln 2 = - 0 .693 ( h ) For step 2 , the process 2 → 3 is isobaric, so heat transfer is, Q = n C p T 3 − T 2 ⇒ Q = n 5 2 R T 2 T 3 T 2 − 1 ( ∵ C p = 5 2 R for a monatomic gas) ⇒ Q = 5 2 n R T 1 T 3 T 2 − 1 ∵ T 2 = T 1 ⇒ Q P 1 V 1 = 5 2 V 3 V 2 − 1 . . . 6 ∵ V 3 T 3 = V 2 T 2 , ⇒ T 3 T 2 = V 3 V 2 We know, process 1 → 2 is isothermal. So, from Boyle's law, ⇒ P 1 V 1 = P 2 V 2 ⇒ P 1 V 1 = 2 P 1 V 2 ⇒ V 2 = V 1 2 Now, V 3 V 2 = 2 V 1 V 1 2 = 4 . . . 7 On putting equation 7 in equation 6 , we get, ⇒ Q P 1 V 1 = 5 2 4 − 1 = 2 5 × 3 = 7 .5 ( i ) For step 1 , process 1 → 2 is isothermal, work done is, W = P 1 V 1 ln V 2 V 1 ⇒ W P 1 V 1 = ln V 1 2 V 1 ∵ V 2 = V 1 2 ⇒ W P 1 V 1 = ln 1 2 = − 0 .693 ( j ) For step 2 , process 2 → 3 is isobaric, so work done is, W = 2 P 1 V 3 − V 2 ⇒ W = 2 P 1 V 2 V 3 V 2 − 1 ⇒ W = 2 P 1 V 1 2 2 V 1 V 1 2 − 1 ∵ V 3 = 2 V 1 and V 2 = V 1 2 ⇒ W = P 1 V 1 4 − 1 ⇒ W P 1 V 1 = 3 .00 ( k ) For the total process, the change in internal energy is, Δ E int = Δ E 1 → 2 + Δ E 2 → 3 ⇒ Δ E 1 → 2 = 0 (Since the process in isothermal) and Δ E 2 → 3 = n C v T 3 − T 2 (Since the process in isobaric) ⇒ Δ E 2 → 3 = n × 3 2 R T 2 T 3 T 2 − 1 ⇒ Δ E 2 → 3 = 3 2 n R T 1 T 3 T 2 − 1 ∵ T 2 = T 1 ⇒ Δ E 2 → 3 P 1 V 1 = 3 2 V 3 V 2 − 1 ∵ P 1 V 1 = n R T 1 and T 3 T 2 = P 3 P 2 ⇒ ∆ E 2 → 3 P 1 V 1 = 3 2 2 V 1 V 1 2 − 1 ∵ V 3 = 2 V 1 and V 2 = V 1 2 ⇒ Δ E 2 → 3 P 1 V 1 = 3 2 4 − 1 = 9 2 = 4 .50 Substituting values of Δ E 1 → 2 and Δ E 2 → 3 for Δ E , we get, ⇒ Δ E int P 1 V 1 = 0 + Δ E 2 → 3 P 1 V 1 ⇒ Δ E int P 1 V 1 = 4 .50 ( l ) For total change in entropy Δ S : We know, in this process, the system moves from state 1 to state 2 . As we know, change in entropy is a point function, it is independent of path followed, so the value of change in entropy in this will be equal to previous case, i.e., Δ S = 23 .051 J K - 1 .

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Earn 100

A gas sample undergoes a reversible isothermal expansion. Figure (as shown below) gives the change $Δ S$ in the entropy of the gas versus the final volume $V_{f}$ of the gas. The scale of the vertical axis is set by $Δ S_{s} = 128 J K^{- 1}$ . How many moles are in the sample?

Important Questions on Entropy and the Second Law of Thermodynamics

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Principles Of Physics International Student Version>Chapter 20 - Entropy and the Second Law of Thermodynamics>Problems>Q 33

A mixture of

1773 g

of water and undefined of ice is in an initial equilibrium state at

0.000 ° C

. The mixture is then, in a reversible process, brought to a second equilibrium state, where the water-ice ratio, by mass, is

1.00 : 1.00

0.000 ° C

(a)

Calculate the entropy change of the system during this process. (Heat of fusion for water is undefined)

(b)

The system is then returned to the initial equilibrium state in an irreversible process (say, by using a Bunsen burner). Calculate the entropy change of the system during this process.

(c)

Are your answers consistent with the second law of thermodynamics?

EASY

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IMPORTANT

Principles Of Physics International Student Version>Chapter 20 - Entropy and the Second Law of Thermodynamics>Problems>Q 34

An ideal gas undergoes a reversible isothermal expansion at

77.0 ° C,

increasing its volume from

1.30 L

3.90 L

. The entropy change of the gas is

22.0 J K^{- 1}

. How many moles of gas are present?

HARD

JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 20 - Entropy and the Second Law of Thermodynamics>Problems>Q 35

Suppose $1.00 mol$ of a monatomic ideal gas is taken from initial pressure $p_{1}$ and volume $V_{1}$ , through two steps: $(1)$ an isothermal expansion to volume $2.00 V_{1}$ and $(2)$ a pressure increase to $2.00 p_{1}$ at constant volume. What is $\frac{Q}{p_{1} V_{1}}$ for $(a)$ step $1$ and $(b)$ step $2$ ? What is $\frac{W}{p_{1} V_{1}}$ for $(c)$ step $1$ and $(d)$ step $2$ ? For the full process, what are $(e)$ $\frac{Δ E_{int}}{p_{1} V_{1}}$ and $(f) Δ S$ ?

The gas is returned to its initial state and again taken to the same final state but now through these two steps:
$(1)$ an isothermal compression to pressure $2.00 p_{1}$ and $(2)$ a volume increase to $2.00 V_{1}$ at constant pressure. What is $\frac{Q}{p_{1} V_{1}}$ for $(g)$ step $1$ and $(h)$ step $2$ ? What is $\frac{W}{p_{1} V_{1}}$ for
$(i)$ step $1$ and $(j)$ step $2$ ? For the full process, what are $(k) \frac{Δ E_{int}}{p_{1} V_{1}}$ and $(l)$ $Δ S$ ?

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Principles Of Physics International Student Version>Chapter 20 - Entropy and the Second Law of Thermodynamics>Problems>Q 36

How much energy must be transferred as heat for a reversible isothermal expansion of an ideal gas at

180 ° C

, if the entropy of the gas increases by

46.0 J K^{- 1}

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JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 20 - Entropy and the Second Law of Thermodynamics>Problems>Q 37

The cycle as shown in the figure represents the operation of a gasoline internal combustion engine. Volume $V_{3} = 4.00 V_{1}$ . Assume the gasoline-air intake mixture is an ideal gas with $γ = 1.30 .$ What are the ratios (a) $\frac{T_{2}}{T_{1}}$ , (b) $\frac{T_{3}}{T_{1}}$ , (c) $\frac{T_{4}}{T_{1}}$ , (d) $\frac{p_{3}}{p_{1}}$ and (e) $\frac{p_{4}}{p_{1}}$ ? (f) What is the engine efficiency?

Question Image

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Principles Of Physics International Student Version>Chapter 20 - Entropy and the Second Law of Thermodynamics>Problems>Q 38

A Carnot engine absorbs

52 kJ

as heat and exhausts

30 kJ

as heat in each cycle. Calculate (a) the engine's efficiency and (b) the work done per cycle in

kJ

HARD

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IMPORTANT

Principles Of Physics International Student Version>Chapter 20 - Entropy and the Second Law of Thermodynamics>Problems>Q 39

In an experiment,

400 g

of aluminium (with a specific heat of

900 J {kg}^{- 1} K^{- 1})

100 ° C

is mixed with

50.0 g

of water at

20.0 ° C

with the mixture thermally isolated. (a) What is the equilibrium temperature? What are the entropy changes of (b) the aluminium, (c) the water, and (d) the aluminium-water system?

HARD

JEE Main

IMPORTANT

Principles Of Physics International Student Version>Chapter 20 - Entropy and the Second Law of Thermodynamics>Problems>Q 40

A two-stage Carnot engine consists of two Carnot cycles: In stage

1,

a Carnot cycle absorbs heat at temperature

T_{1} = 500 K

and discharges heat at temperature

T_{2} = 400 K

. In stage

2,

a Carnot cycle absorbs that discharged heat at

T_{2}

and then discharges heat at temperature

T_{3} = 300 K .

What is the efficiency of the engine?

A gas sample undergoes a reversible isothermal expansion. Figure (as shown below) gives the change ΔS in the entropy of the gas versus the final volume Vf of the gas. The scale of the vertical axis is set by ΔSs=128 J K-1. How many moles are in the sample?

Important Questions on Entropy and the Second Law of Thermodynamics

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A gas sample undergoes a reversible isothermal expansion. Figure (as shown below) gives the change $Δ S$ in the entropy of the gas versus the final volume $V_{f}$ of the gas. The scale of the vertical axis is set by $Δ S_{s} = 128 J K^{- 1}$ . How many moles are in the sample?