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Thermodynamics - 1 - JEE MCQ

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30 Questions MCQ Test Chemistry for JEE Main & Advanced - Thermodynamics - 1

Thermodynamics - 1 for JEE 2024 is part of Chemistry for JEE Main & Advanced preparation. The Thermodynamics - 1 questions and answers have been prepared according to the JEE exam syllabus.The Thermodynamics - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Thermodynamics - 1 below.

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Thermodynamics - 1 - Question 1

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The pressure-volume work for an ideal gas can be calculated using the expression

This type of work can also be calculated using the area under the curve within the specified limits. When an ideal gas is compressed, (I) reversibly or (II) irreversibly, then

A.
W(I) = W(II)
B.
W(I) < W(II)
C.
W(I) > W(II)
D.
W(I) = W (II) + p_exΔV

Detailed Solution for Thermodynamics - 1 - Question 1

Work done is the area under the P−V curve. It can be seen in the curve above that the area under the P−V curve for irreversible compression of the gas is more than the area under the curve for reversible compression.
Thus, work done for irreversible compression is more than that for reversible compression.

Thermodynamics - 1 - Question 2

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In the following case

expanded in reversible adiabatic condition to make volume 8 times ( γ = 1.33). Find the Work Done and Temperature.

A.
a
B.
b
C.
c
D.
d

Detailed Solution for Thermodynamics - 1 - Question 2

As q = 0, we have adiabatic process.
V₁ = 100L and V₂ = 800L
T₁ = 300K and T₂ = not given
For NH₃, γ = 4/3
Applying TV^γ = constant
(300)(100)^4/3-1 = (T)(800)^4/3-1
T = 300/2 = 150K
W= nR(T₂-T₁)/γ-1
= 1×8.314×150/(4/3-1)
= 3714 J
= 900 cal

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Thermodynamics - 1 - Question 3

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1.0 mole of a monoatomic ideal gas is expanded from state I to state II at 300 K.

Thus, work done is

A.
-1724 J
B.
+1729 J
C.
-75.8 J
D.
17.07 J

Detailed Solution for Thermodynamics - 1 - Question 3

Work done in isothermal process
W= -2.303 nRT log P1/P2
here n=1 and R =8.314 and T= 300 P1=4and P2=2.
W= -1724 J
So option A is correct.

Thermodynamics - 1 - Question 4

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1 mole of a diatomic gas is contained in a piston. It gains 50.0 J of heat and work is done on the surrounding by the system is -100 J. Thus,

A.
the gas will cool by 2.41°
B.
he gas will heat by 2.41°
C.
the gas will cool by 3.61°
D.
he gas will heat by 3.61°

*Multiple options can be correct

Thermodynamics - 1 - Question 5

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A sample containing 1.0 mole of an ideal gas is expanded isothermally and reversibly to ten time of its original volume, in two separate experiments. The expansion is carried out 300 K and at 600 K, respectively. Choose the correct option.

A.
Work done at 600 K is 20 times the work done at 300 K
B.
Work done at 300 K is twice the work done at 600 K
C.
Work done at 600 K is twice the work done at 300 K
D.
ΔE = 0 in both cases

Detailed Solution for Thermodynamics - 1 - Question 5

Work done in isothermal process, w = nRTln(V₂/V₁)
w₆₀₀/w₃₀₀ = 1×R×600×ln(10)/ 1×R×300ln(10) = 2
∆U = 0 for isothermal processes.

Thermodynamics - 1 - Question 6

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A.
–196 kJ/mol
B.
+196 kJ/mol
C.
+948 kJ/mol
D.
–948 kJ/mol

Detailed Solution for Thermodynamics - 1 - Question 6

Thermodynamics - 1 - Question 7

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Detailed Solution for Thermodynamics - 1 - Question 7

Thermodynamics - 1 - Question 8

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During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. The change in internal energy, ΔU (in kJ) is :

A.
12
B.
–8
C.
8
D.
–12

Detailed Solution for Thermodynamics - 1 - Question 8

Work done on system = +10 kJ
Heat escaped = –2kI
ΔU = q + w
= 10 – 2 = 8 KJ.

Thermodynamics - 1 - Question 9

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A.
11.4 kJ
B.
–11.4 kJ
C.
0 kJ
D.
4.8 kJ

Detailed Solution for Thermodynamics - 1 - Question 9

Thermodynamics - 1 - Question 10

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A.
38
B.
317
C.
477
D.
32

Detailed Solution for Thermodynamics - 1 - Question 10

Thermodynamics - 1 - Question 11

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Detailed Solution for Thermodynamics - 1 - Question 11

Thermodynamics - 1 - Question 12

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Detailed Solution for Thermodynamics - 1 - Question 12

Thermodynamics - 1 - Question 13

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Detailed Solution for Thermodynamics - 1 - Question 13

Thermodynamics - 1 - Question 14

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A.
910 K
B.
1110 K
C.
510 K
D.
710 K

Detailed Solution for Thermodynamics - 1 - Question 14

Thermodynamics - 1 - Question 15

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Detailed Solution for Thermodynamics - 1 - Question 15

Thermodynamics - 1 - Question 16

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A.
+RT
B.
–3RT
C.
+3RT
D.
–RT

Detailed Solution for Thermodynamics - 1 - Question 16

Thermodynamics - 1 - Question 17

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A.
10.52 cal/(mol K)
B.
21.04 cal/(mol K)
C.
5.260 cal/(mol K)
D.
0.526 cal/(mol K)

Detailed Solution for Thermodynamics - 1 - Question 17

Thermodynamics - 1 - Question 18

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A.
Br₂ (g)
B.
Cl₂ (g)
C.
H₂O (g)
D.
CH₄ (g)

Detailed Solution for Thermodynamics - 1 - Question 18

Thermodynamics - 1 - Question 19

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A.
negative
B.
positive
C.
zero
D.
constant

Detailed Solution for Thermodynamics - 1 - Question 19

Thermodynamics - 1 - Question 20

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Detailed Solution for Thermodynamics - 1 - Question 20

Thermodynamics - 1 - Question 21

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Detailed Solution for Thermodynamics - 1 - Question 21

Thermodynamics - 1 - Question 22

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A.
–17 kcal
B.
–111 kcal
C.
–170 kcal
D.
–85 kcal

Detailed Solution for Thermodynamics - 1 - Question 22

Thermodynamics - 1 - Question 23

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For silver, C_p(JK^–1 mol^–1) = 23 + 0.01 T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of ΔH will be close to -

A.
13 kJ
B.
16 kJ
C.
62 kJ
D.
21 kJ

Detailed Solution for Thermodynamics - 1 - Question 23

Thermodynamics - 1 - Question 24

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Detailed Solution for Thermodynamics - 1 - Question 24

Thermodynamics - 1 - Question 25

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A.
525 kJ/mol
B.
-175 kJ/mol
C.
-325 kJ/mol
D.
325 kJ/mol

Detailed Solution for Thermodynamics - 1 - Question 25

Thermodynamics - 1 - Question 26

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Detailed Solution for Thermodynamics - 1 - Question 26

Thermodynamics - 1 - Question 27

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A.
273.4 K
B.
393.4 K
C.
373.4 K
D.
293.4 K

Detailed Solution for Thermodynamics - 1 - Question 27

Thermodynamics - 1 - Question 28

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Detailed Solution for Thermodynamics - 1 - Question 28

Thermodynamics - 1 - Question 29

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A.
1118 K
B.
1008 K
C.
1200 K
D.
845 K

Detailed Solution for Thermodynamics - 1 - Question 29

Thermodynamics - 1 - Question 30

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A.
92.1
B.
0
C.
4
D.
9.2

Detailed Solution for Thermodynamics - 1 - Question 30

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Thermodynamics - 1 - JEE MCQ

30 Questions MCQ Test Chemistry for JEE Main & Advanced - Thermodynamics - 1

The pressure-volume work for an ideal gas can be calculated using the expression This type of work can also be calculated using the area under the curve within the specified limits. When an ideal gas is compressed, (I) reversibly or (II) irreversibly, then

In the following case expanded in reversible adiabatic condition to make volume 8 times ( γ = 1.33). Find the Work Done and Temperature.

1.0 mole of a monoatomic ideal gas is expanded from state I to state II at 300 K. Thus, work done is

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Important Questions for Thermodynamics - 1

Thermodynamics - 1 MCQs with Answers

Online Tests for Thermodynamics - 1 Chemistry for JEE Main & Advanced

The pressure-volume work for an ideal gas can be calculated using the expression

This type of work can also be calculated using the area under the curve within the specified limits. When an ideal gas is compressed, (I) reversibly or (II) irreversibly, then

In the following case

expanded in reversible adiabatic condition to make volume 8 times ( γ = 1.33). Find the Work Done and Temperature.

1.0 mole of a monoatomic ideal gas is expanded from state I to state II at 300 K.

Thus, work done is