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Test: First law, Thermodynamic Equilibrium - JEE MCQ


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15 Questions MCQ Test Chemistry for JEE Main & Advanced - Test: First law, Thermodynamic Equilibrium

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

A gas expands adiabatically at constant pressure such that . The value of of the gas will be :

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 1


For adiabatic process, constant

Test: First law, Thermodynamic Equilibrium - Question 2

What is the value of change in internal energy at 1 atm in the process?

Given :

at

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 2

Given,

The steps involved in the above conversion is

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Test: First law, Thermodynamic Equilibrium - Question 3

An ideal gas expands against a constant external pressure of 2.0 atmosphere from 20 litre to 40 litre and absorbs 10 kJ of heat from surrounding. What is the change in internal energy of the system? (given : 1 atm-litre = 101.3 J)

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 3


Test: First law, Thermodynamic Equilibrium - Question 4

The pressure-volume diagram shows six curved paths that can be followed by the gas (connected by vertical paths). Which two of them should be part of a closed cycle if the net work done by the gas is to be its maximum positive value?

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 4

Work done in expansion is positive and in compression it is negative.

Test: First law, Thermodynamic Equilibrium - Question 5

On P − V coordinates, the slope of an isothermal curve of a gas at a pressure P = 1MPa and volume V = 0.0025 m3 is equal to −400MPa/m3. If Cp/Cv = 1.4, the slope of the adiabatic curve passing through this point is :

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 5

Slope of adiabatic curve = γ × slope of isothermal curve = 1.4 × (−400) = −560MPa/m3

Test: First law, Thermodynamic Equilibrium - Question 6

For an ideal gas four processes are marked as 1 2,3 and 4 on diagram as shown in figure. The amount of heat supplied to the gas in the process and 4 are and respectively, then correct order of heat supplied to the gas is is process- is process- is adiabatic process-3 and is process-4]

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 6

In process- 1 heat supplied = area under AB curve +n × cp × 100 (isochoric process) In process-2 heat supplied = area under AC curve (isothermal process) In process-3 heat supplied =0 (adiabatic process) In process- 4 heat supplied = n × cv(T − 600)
(isobaric process)

Test: First law, Thermodynamic Equilibrium - Question 7
Which one of the following is applicable for an adiabatic expansion of an ideal gas?
Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 7

For an adiabatic reaction,

So,
Test: First law, Thermodynamic Equilibrium - Question 8

A system changes from the state to ) as shown in the figure. What is the work done by the system?

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 8



Joule

Test: First law, Thermodynamic Equilibrium - Question 9

If 2.5 moles of an ideal gas at a certain temperature are allowed to expand isothermally and reversibly from an initial volume of to , the work done by the gas is −16.5 kJ.
The temperature (in K) of the gas is (Round off to the nearest value)

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 9

Reversible isothermal work is given by :

Test: First law, Thermodynamic Equilibrium - Question 10

An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states



The ratio of the net work done by the gas to the total heat absorbed by the gas is . The values of and respectively are

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 10





In cyclic process, Now,
or


Test: First law, Thermodynamic Equilibrium - Question 11

For a monoatomic gas, the value of the ratio of and is

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 11

For a monoatomic gas, CV, m = (3/2)R and Cp,m = (5/2)R . Hence Cp,m/CV,m = 5/3. 

Test: First law, Thermodynamic Equilibrium - Question 12

If is the change in enthalpy and the change in internal energy accompanying a gaseous reaction, then

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 12

As
If
Hence

Test: First law, Thermodynamic Equilibrium - Question 13

An ideal gas expands adiabatically against vaccum. Which of the following is correct for the given process?

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 13

Hint:
For adiabatic free expansion of an ideal gas



Test: First law, Thermodynamic Equilibrium - Question 14

When the state of a gas adiabatically changed from an equilibrium state A to another equilibrium state B an amount of work done on the stystem is . If the gas is taken from state to via process in which the net heat absorbed by the system is , then the net work done by the system is

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 14

In the first-case adiabatic change,
From law of thermodynamics,
or

In the second case

Test: First law, Thermodynamic Equilibrium - Question 15

The ratio γ = Cp, m/CV, m for a diatomic gaseous molecules is

Detailed Solution for Test: First law, Thermodynamic Equilibrium - Question 15

To find:
Cp/C for diatomic gas.
Cp → Molar specific heat capacity at constant pressure.
Cv → Molar specific heat capacity at constant volume..

f → degree of freedom of gas molecules.
A molecule of diatomic gas has 5 degree of freedom, i.e., f = 5.

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