The molar heat capacity, C v of helium gas is 3 R 2 and is independent of temperature. For hydrogen gas, C v approaches 3 R 2 at very low temperatures, equal 5 R 2 at moderate temperatures and is higher than 5 R 2 at high temperatures. Give a reason for the temperature dependence of C v in case of hydrogen, in not more than two or three sentences. [ JEE 2003 , 2 / 60 ]

Helium is mono atomic gas, so it have only translational motion.It can move in any direction in space (along X,Y,Z axis), hence it has 3 degree of freedom. The contribution of each degree of freedom towards C v is R 2 , so total value of C v by three degree of freedom is 3 2 R . For diatomic molecules like H 2 at very low temperature there is only translational motion, hence it also has degree of freedom 3 . So at very low temperature total value of C v is 3 2 R . At moderate temperature diatomic gas has 3 translational as well as 2 rotational degree of freedom, so total value of C v is 5 2 R . At very high temperature vibrational motion also counts hence total degree of freedom for diatomic gas increased as well as total value of C v .

One mole of ideal monatomic gas was taken through isochoric heating from 100 K to 1000 K Calculate Δ S system , Δ S surr and Δ S total in. when the process is carried out reversibly?

An isochoric process is a constant volume process. Since entropy is a point functions, change of entropy between two points would be some irrespective of the process being reversible or irreversible or of different path. ∴ △ S sys irrev . onestep = △ S sys rev = ∫ 1 2 dQ T △ Ssys = ∫ 100 1000 nCrdT T Given n = 1 , Cr = 3 R monoatomic ∴ △ Ssys = 3 2 R [ lnT ] 100 1000 = 3 2 Rln 10 Now, △ S sys + △ S surrouding = △ S Total > 0 for irreversible process = - 3 2 R 1000 − 100 1000 = − 3 2 R 0 . 9 = 3 2 R 1 . 403

One mole of ideal monoatomic gas was taken through isochoric heating from 100 K to 1000 K Calculate Δ S system , Δ S surr and Δ S total in. when the process carried out irreversibly (one step)

Change of entropy between two points would be some irrespective of the process being reversible or irreversible or of different path because entropy is a point function. ∴ ( △ S sys ) irrev . = ( △ S sys ) rev = ∫ T 1 T 2 dQ T △ S sys = ∫ 100 1000 nC v dT T This is a mono atomic gas. So, it has the C v = 3 2 R So, △ S sys = 3 2 R [ lnT ] 100 1000 = 3 2 R ln 1000 - ln 100 = 3 2 Rln 10 Now, ( △ S ) sys + ( △ S ) surrouding = ( △ S ) Total > 0 for irreversible process ( △ S ) surrouding = - ∆ Q T surrouding = - nC v ∆ T T surrouding = - 3 2 R ( 1000 − 100 ) 1000 = − 3 2 R ( 0 . 9 ) ( △ S ) Total = ( △ S ) sys + ( △ S ) surrouding = 3 2 Rln 10 + − 3 2 R ( 0 . 9 ) = 3 2 R ( ln 10 - 0 . 9 ) = 3 2 R ( 1 . 403 ) Hence, Irr. Process ΔS system = 3 2 Rln 10 ; ΔS sur = - 3 2 R ( 0 . 9 ) ; ΔS total = 3 2 R ( 1 . 403 )

State the second law of thermodynamics.

The second law of thermodynamics can be expressed in terms of entropy. It can be stated as “the entropy of an isolated system increases during a spontaneous process”. The total entropy change ( ∆ S total ) for the system and the surroundings of a spontaneous process is given by, ∆ S total = ∆ S system + ∆ S surrounding > 0 When a system is in equilibrium, the entropy is maximum and the change in entropy, ∆ S = 0 . Hence, entropy of a spontaneous process increases till it reaches a maximum and at equilibrium the change in entropy is zero.

The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at 25 ° C is

298 × 10 7 × 8 . 31 × 2 . 303 log 2 .

2 . 303 × 298 × 0 . 082 log 2 .

2 . 303 × 298 × 0 . 082 log 0 . 5 .

8 . 31 × 10 7 × 298 × 2 . 303 log 0 . 5 .

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by:

with increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases.

with increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive.

with increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative.

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Hydrazine used in rocket fuels can be obtained by the reaction of ammonia and hydrogen peroxide according to the following equations

$2 {NH}_{3} (g) + H_{2} O_{2} (I) ⟶ N_{2} H_{4} (I) + 2 H_{2} O (I) (Δ H_{reaction} ° = - 241 kJ / mol)$

If $Δ H °$ (formation) of ${NH}_{3}, H_{2} O_{2}$ and $H_{2} O$ are $- 46.1, - 187.8$ and $- 285.8 kJ / mol$ respectively, $Δ H °$ for the decomposition of hydrazine into $N_{2}$ and $H_{2}$ is

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Important Questions on Thermodynamics

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 6

One mole of Ideal gas $(C_{P, m} = 15 {JK}^{- 1} {mole}^{- 1})$ . follow the process as shown in figure. Predict the following:

Question Image

Calculate work done, Heat of process.

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 7

1

mole of

{CO}_{2}

300 K

is expanded under reversible adiabatic condition such that its volume becomes

27

times.
(a) What is the final temperature?
(b) What is work done? Given

γ = 1.33

and

C_{V} = 25.08 Jmol {}^{- 1}K^{- 1}

for

{CO}_{2}

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 8

A certain mass of a gas initially at $(1 L, 5 atm, 300 K)$ is expanded reversibly and isothermally to a final volume of $5 L,$

(a) calculate work done by the gas and heat supplied in this process to the gas.

(b) Now, if the gas is restored to initial position by compressing it using an external constant pressure of $5$ atm. Find work done on the gas in this process and heat rejected by gas
(c) the above two processes, what is the net heat gained by surroundings?

[Note: From above question see that surroundings has done extra work on the system but system has returned that work in the form of heat to surroundings and work is considered on organised form of energy while heat as an unorganised form hence in the above process, there must be net increment in randomness of universe which will be called Entropy, soon.]

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 9

The molar heat capacity,

C_{v}

of helium gas is

\frac{3 R}{2}

and is independent of temperature. For hydrogen gas,

C_{v}

approaches

\frac{3 R}{2}

at very low temperatures, equal

\frac{5 R}{2}

at moderate temperatures and is higher than

\frac{5 R}{2}

at high temperatures. Give a reason for the temperature dependence of

C_{v}

in case of hydrogen, in not more than two or three sentences.

[JEE 2003, 2 / 60]

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 11

One mole of monoatomic gas was taken through a cyclic process as shown in the figure. Calculate

$\sum_{ABCA} \frac{{dq}_{rev}}{T}$

Question Image

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 12

One mole of ideal monatomic gas was taken through isochoric heating from

100 K

1000 K

Calculate

Δ S_{system}, Δ S_{surr}

and

Δ S_{total}

in.
when the process is carried out reversibly?

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 13

One mole of ideal monoatomic gas was taken through isochoric heating from

100 K

1000 K

Calculate

Δ S_{system}, Δ S_{surr}

and

Δ S_{total}

in.
when the process carried out irreversibly (one step)

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Alpha Question Bank for Engineering: Chemistry>Chapter 4 - Thermodynamics>Exercise-5>Q 14

Comment on the following statements:
An exothermic reaction is always thermodynamically spontaneous.

Important Questions on Thermodynamics

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