Thermodynamics Test

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Thermodynamics Test - 6

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Weekly Quiz Competition

Question 1
4 / -1

Specific heat of iron at 25°C and 1.00 bar is

A
0.4454 JK^-1 g^-1

B
24.942 JK^-1 mol^-1

C
0.1071 cal K^-1 g^-1

D
All are correct
Question 2
4 / -1

A rather soft, silvery metal was observed to have a specific heat of 0.225 cal K^-1g^-1. Thus, metal is

A
Na

B
Mg

C
Al

D
Fe

Solution

It should be better to remove this question. It's data based.
Question 3
4 / -1

Specific heat of water is 4.184 JK^-1g^-1, Rise in temperature when 1 kJ of heat is absorbed by 2 moles of H₂O is

A
119.5^o

B
0.37^o

C
0.12^o

D
6.64^o

Solution

We have,
Q = ms∆T
1000 = 2*18*4.184*∆T
∆T = 6.63 K
Question 4
4 / -1

The amount of the heat released when 20 ml 0.5 M NaOH is mixed with 100 ml 0.1 M HCl is x kJ. The heat of neutralization is

A
– 100 x kJ/mol

B
– 50 x kJ/mol

C
+ 100 x kJ/mol

D
+50 x kJ/mol

Solution
Question 5
4 / -1

The heat capacity ratio, r was determined for cyanogen as 1.177. Thus, C_p for this gas is

A
55.3 JK^-1 mol^-1

B
33.26 JK^-1 mol^-1

C
8.314 JK^-1 mol^-1

D
24.94 JK^-1 mol^-1

Solution

γ = 1.177
C_V = R/γ-1
= 8.314/0.177 = 46.97
Also γ = C_P/C_V
1.177 = C_P/46.97
Or C_P= 55.28
Question 6
4 / -1

Assuming the composition of air to be X (N₂) = 0.80, X (O₂) = 0.18 and X (CO₂) = 0.02, molar heat capacity of air at constant pressure is

A
3.50R

B
4.00R

C
2.50R

D
3.51R

Solution

(C_p)_mixture = (µ₁C_p1+µ₂C_p2+µ₃C_p3+.........)/µ₁+µ₂+µ₃+......... where µ is the no of moles.
Or (C_p)_mixture = ƞ₁C_p1+ƞ₂C_p2+ƞ₃C_p3+............. Where ƞ is the corr. mole fraction.
On putting the values, we have
Cp_mixture= ( 7/2*0.8 +7/2*0.18 +4*0.02)R
=3.15R
Question 7
4 / -1

Latent heat of fusion of ice is 6.02 kJ mol^-1. The heat capacity of water is 4.18 Jg^-1K^-1. 500 g of liquid water is to be cooled from 20°C to 0°C . Number of ice cubes (each of one mole) required is

A
1

B
7

C
14

D
125

Solution

Molecular mass of hydrogen = 1
Molecular mass of oxygen =16
Molecular mass of water = 18 gram per mole
As,
n = m / M
n = 500 / 18
n = 27.8 moles
For cooling 27.8 moles at 20 = 27.8 x 75.4 x 20 = 41.922 kilo joule
41.922/6 = 6.987 that is approximately equal to 7
So, the number of ice cubes that are required to cool the water will be 7.
Question 8
4 / -1

Sulphur (2.56 g) is burned in a constant volume calorimeter with excess O₂(g). The temperature increases from 21.25°C to 26.72°C . The bomb has a heat capacity of 923 JK^-1. Calorimeter contains 815 g of water. Thus, change in internal energy per mole of SO₂ formed for the reaction is
(specific heat of water is 4.184 JK^-1g^-1.)

A
- 296.27 kJ

B
+ 296.27 kJ

C
- 2370.13 kJ

D
+ 2370.13 kJ

Solution

Moles of S8 = 2.56/256 = 0.01
So, moles of SO₂ formed = 0.08
Rise in temperature = (273+26.72)K - (273+21.25)K
= 5.47 K
Total Energy in internal energy (as system is at constant volume) = -(923 jK-1 × 5.47 +815g×4.18)
= -23701.29 J = -23.70129 kJ
Thus change in internal energy per mole of formed = -23.7012kJ/0.08 = -296.3 kJ
Question 9
4 / -1

This sectionis based on statement I and Statement II. Select the correct answer from the code given below.

Q.

Statement I : C_v value of helium (He) is always 3/2R but C_v value of hydrogen (H₂) is 3/2R at low temperature, 5/2R at moderate temperature and more than 5/2R at higher temperature.

Statement II : At lower temperature, only translational degree of freedom contributes to heat capacity while at higher temperature, rotational and vibrational degrees of freedom also contribute to heat capacity.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

In case of helium (monoatomic gas) we have only three degrees of freedom which correspond to three translational motion so the total heat capacity will increase. The contributors by vibrational motion are not appreciable at low temperature but increase from 0 to R when temperature increases.
CV=−f(R)/2
CV=-f(R)2, where f is the degree of freedom. At low temperature only translational motion is considered and f=3
∴CV=3R/2
f=3.∴CV=3R2 At moderate temperature both translational and rotational motions are considered.
f=3+2
f=3+2 (3-translational and 2 rotational).
∴CV=5R/2
∴CV=5R2. At still high temperature translational, rotational and vibrational motions are considered.
f=3+2+2
f=3+2+2 (3-translatinal , 2-rotational, 2-vibrational).
∴CV=7R/2
Question 10
4 / -1

On which of the following factors the enthalpy change of a reaction is NOT dependent?

A
Nature of reactants and products

B
Various intermediate reactions

C
State of reactants and products

D
Enthalpy of reactants and products

Solution

Enthalpy change for any process is a state function. It depends only on the initial and final conditions.
Similarly, Enthalpy change of a reaction does not depend upon different intermediate reactions.
△H=H_p−H_r
Question 11
4 / -1

In an endothermic reaction, the value of ΔH is always

A
Constant

B
< 0

C
> 0

D
= 0

Solution

According to me, for an endothermic reaction ,the value of ∆H is always >0.
Question 12
4 / -1

Direction (Q. Nos. 12 - 14) This section contains 3 questions. when worked out will result in an integer from 0 to 9 (both inclusive).

Q. An 88.5 g piece of iron whose temperature is 352.0 K is placed in a beaker containing 244 g of water at 292.0 K. When thermal equilibrium is reached, what is the change in the temperature of water?
Specific heat of water and iron are (4.184 Jg^-1 K^-1) and (0.449 Jg^-1 K^-1), respectively.

A
2

B
3

C
4

D
5

Solution

Heat lost by iron = Heat gained by water
Let the equilibrium temperature be T.
m_Irons_Iron (352-T) = m_waters_water(T-292)
88.5×0.449×(352-T) = 244×4.18×(T-292)
On calculation approximation,
40(352-T) = 1020(T-292)
0.04(352-T) = (T-292)
T = 306/1.04 = 294 K
So, change in temperature of water = 294-292 = 2K
Question 13
4 / -1

The heat capacity of a calorimeter (commonly called the calorimeter constant) was determined by heating the calorimeter and its content using an electrical heater. If ΔT = 1.221 K as 1.25 A of electricity at 3.26 V was passed through the heater immersed in 137.5 g of water in the calorimeter for 175 s, determine the calorimeter constant (in JK^-1). (Specific heat of water at constant pressure is 75.291 JK^-1 mol^-1).

A
7

B
5

C
6

D
9

Solution

Subtracting the energy gained by the cold water from the energy lost by the hot water. This will give us the amount of energy gained by the calorimeter. Dividing the energy gained by the calorimeter by Tc (the temperature change of the cold water). This final answer is calorimeter constant
ms(dT)=137.5×4.184×1.221=702
joules=coulomb×volt =. 1.25×175×3.26=713
713 - 702 = 11
(11/1.221) = 9
Question 14
4 / -1

In a constant volume calorimeter, 3.5 g of a gas (molar mass = 28 g mol^-1) was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to oxidation of the gas. Given that the heat capacity of the calorimeter is 2.5 kJ K^-1, the numerical value of the heat given out in kJ mol^-1 is

[IIT JEE 2009]

A
9

B
8

C
6

D
3

Solution

Energy released by combustion of 3.5 g gas = 2.5×(298.45-298) kJ
Energy released by 1 mole of gas = 2.5×0.45/3.5/28 = 9 kJ mol^-1
Question 15
4 / -1

Boiling point of ethanol (C₂H₅OH) = 78.29°C
Specific heat of ethanol = 2.44 J^-1K^-1
Enthalpy of vaporisation = 855 Jg^-1

Q. Energy required (in kJ) for the following change is

A
997.23 KJ

B
142.23 KJ

C
3.092 KJ

D
21.68 KJ

Solution

Q = ms∆T
= 1000(2.44×58.29)
= 997.23 kJ
As we have been asked till boiling point. But the compound is still in a liquid state. So mL is not given
Question 16
4 / -1

Boiling point of ethanol (C₂H₅OH) = 78.29°C
Specific heat of ethanol = 2.44 J^-1K^-1
Enthalpy of vaporisation = 855 Jg^-1

Q. Energy required (in kJ) for the following change is

A
422.23 KJ

B
997.23 KJ

C
855.0 KJ

D
1139.46 KJ

Solution

As liquid is already at boiling point initially and after giving heat it still is at boiling point. It means that only latent heat is given.
Q = mL = 1000×855 = 855 kJ
Question 17
4 / -1

Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.

Q. According to the classical equipartition of energy theory, the contributions to the molar heat capacity at constant volume are for each degree of translational freedom, for each degree of rotational freedom and (R) for each degree of translational freedom. Match the species in Column I with the values of molar heat capacity (C_p or C_V) in Column II and select answer from codes given below the table.

A
a

B
b

C
c

D
d

Solution

For monoatomic gas, C_P = 5R/2
For diatomic gas, C_P = 9R/2
For triatomic gas, C_V = 6R and C_P = C_V+R = 7R