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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 2025 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 2025 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

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

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

In the following case, find the option with the correct matching

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 TVγ = 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

Thus,  T2 =  150K and W = 900 cal
Thus option A is correct

Thermodynamics - 1 - Question 3

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

Thus, work done is

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= -1729 J 
So option A is correct.

Thermodynamics - 1 - Question 4

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,

Detailed Solution for Thermodynamics - 1 - Question 4
- To determine the change in temperature of the diatomic gas, we can use the first law of thermodynamics: ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added, and W is the work done by the gas.
- Here, Q = 50.0 J and W = -100 J (since work is done on the surroundings, it is negative).
- Therefore, ΔU = 50.0 J - (-100 J) = 150 J.
- For 1 mole of a diatomic gas, ΔU = (5/2) n R ΔT. Using R = 8.31 J/(mol·K), we find ΔT = ΔU / ((5/2) n R) = 150 J / ((5/2) * 1 * 8.31) = 3.61 K.
- Since the internal energy increased, the gas heats up by 3.61°C.
*Multiple options can be correct
Thermodynamics - 1 - Question 5

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.

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

Change in enthalpy for reaction,
2H2O2(l) → 2H2O(l) + O2(g)
if heat of formation of H2O2(l) and H2O(l) are −188 and −286 kJ/mol respectively, is

Detailed Solution for Thermodynamics - 1 - Question 6

Thermodynamics - 1 - Question 7

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

Detailed Solution for Thermodynamics - 1 - Question 7

Thermodynamics - 1 - Question 8

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 :
 

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

Two moles of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change (in kJ) for the process is

Detailed Solution for Thermodynamics - 1 - Question 9

ΔH = nCp ΔT solution; since ΔT = 0 so, ΔH = 0

Thermodynamics - 1 - Question 10

When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0°C and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ (ΔH_comb. (CH₄) = 890 kJ mol⁻¹, ΔH_comb. (C₃H₈) = 2220 kJ mol⁻¹) is

Detailed Solution for Thermodynamics - 1 - Question 10

Thermodynamics - 1 - Question 11

Consider the following reactions:

Enthalpy of formaton of H2O(l) 

Detailed Solution for Thermodynamics - 1 - Question 11

(d): The amount of heat absorbed or released when 1 mole of a substance is directly obtained from its constituent elements is called the heat of formation or enthalpy of formation.
Equation (i) represents neutralisation reaction, (iii) represents hydrogenation reaction and (iv) represents combustion reaction.

Thermodynamics - 1 - Question 12

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant 
Kc  is

Detailed Solution for Thermodynamics - 1 - Question 12

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Thermodynamics - 1 - Question 23

For silver, Cp(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 -

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