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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A heat pump is used to heat a building. The external temperature is less than the internal temperature. The pump's coefficient of performance is $3.30$ and the heat pump delivers $7.54 MJ$ as heat to the building each hour. If the heat pump is a Carnot engine working in reverse, at what rate must work be done to run it?

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 30

A Carnot engine is set up to produce a certain $W$ work per cycle. In each cycle, energy in the form of heat $Q_{H}$ is transferred to the working substance of the engine from the higher-temperature thermal reservoir, which is at an adjustable temperature $T_{H}$ . The lower-temperature thermal reservoir is maintained at temperature $T_{L} = 250 K .$ Given figure gives $Q_{H}$ for a range of $T_{H} .$ The scale of the vertical axis is set by $Q_{H,} = 12.0 kJ .$ If $T_{H}$ is set at $550 K$ , what is $Q_{H} ?$

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Principles Of Physics International Student Version>Chapter 20 - Entropy and the Second Law of Thermodynamics>Problems>Q 31

Find (a) the energy absorbed as heat and (b) the change in entropy of a

3.30 kg

block of copper whose temperature is increased reversibly from

25.0 ° C

170 ° C

. The specific heat of copper is

386 J {kg}^{- 1} K^{- 1}

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Principles Of Physics International Student Version>Chapter 20 - Entropy and the Second Law of Thermodynamics>Problems>Q 32

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?

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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?

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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?

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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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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?

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Important Questions on Entropy and the Second Law of Thermodynamics

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