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Test: First Law of Thermodynamics - NEET MCQ


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20 Questions MCQ Test - Test: First Law of Thermodynamics

Test: First Law of Thermodynamics for NEET 2024 is part of NEET preparation. The Test: First Law of Thermodynamics questions and answers have been prepared according to the NEET exam syllabus.The Test: First Law of Thermodynamics MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: First Law of Thermodynamics below.
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Test: First Law of Thermodynamics - Question 1
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Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q. A gas is cooled and loses 65 J of heat. The gas contracts as it cools and work done on the system equal to 22 J is exchanged with the surroundings. Thus, q, W and ΔE (change in internal energy) are (in joules)

       

Detailed Solution for Test: First Law of Thermodynamics - Question 1

The IUPAC convention is as follow:-
For expansion or work done by the system on surrounding, W = negative
For compression or work done on system by surrounding, W = positive
Heat absorbed by the system is positive and heat released by the system is negative.
So, we have q = -65 J and W = 22 J
From 1st law of thermodynamics,
      ∆U = q+W
On putting values, we get ∆U = -43 J
So, q = -65 J, W = 22 J and ∆U = -43 J

Test: First Law of Thermodynamics - Question 2
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Which of the following is true for a steady flow system?

Detailed Solution for Test: First Law of Thermodynamics - Question 2

The amount of mass or energy entering a control volume does not have to be equal to the amount of mass or energy leaving during an unsteady-flow process.It is mostly converted to internal energy as shown by a rise in the fluid temperature.

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Test: First Law of Thermodynamics - Question 3
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One kg of carbon produces __________ kg of carbon dioxide.

Detailed Solution for Test: First Law of Thermodynamics - Question 3

2CO + O2 ------------> 2CO2
Atomic wt of C =12 kg or gms, Atomic wt of O = 16 kg or gms
2(12 + 16) kg of CO + 32 kg of O2 -----------> 2(12+32) kg of CO2
56 kg of CO + 32 kg of O2 -------> 88 kg of CO2
For, 1kg of CO + 32/56 kg of O2 ---------> 88/56 kg of CO2
Therefore, 1kg of CO requires 32/56 or 4/7 kg of oxygen to produce 88/56kg or 11/7 kg of CO2

Test: First Law of Thermodynamics - Question 4
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A gas absorbs 200 J of heat and expands by 500 cm3 against a constant pressure of 2 x 105 Nm-2. Change in internal energy is 
Detailed Solution for Test: First Law of Thermodynamics - Question 4

The expression for the change in internal energy is given by,
ΔU=q+w
q = heat absorbed = 200 J
w = work done = −P x V
=−2×10⁵ × 500×10−6
= −100 N - m
ΔU=200−100J= 100 J
Hence, the correct option is C.

Test: First Law of Thermodynamics - Question 5
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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
Detailed Solution for Test: First Law of Thermodynamics - Question 5

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.

Test: First Law of Thermodynamics - Question 6
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In the following case

Magnitude of work done and Final temperature respectively are:
Detailed Solution for Test: First Law of Thermodynamics - Question 6

The given images indicate that options (a) and (c) are correct 

Test: First Law of Thermodynamics - Question 7
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Assuming that water vapour is an ideal gas, internal energy change (ΔE)when 1 mole of water is vaporised at 1 bar pressure and 100°C will be

(Given, molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol-1)

[IIT JEE 2007]
Detailed Solution for Test: First Law of Thermodynamics - Question 7

Test: First Law of Thermodynamics - Question 8
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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
Detailed Solution for Test: First Law of Thermodynamics - Question 8

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.

Test: First Law of Thermodynamics - Question 9
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A " 1/4 HP" electric motor uses 187 W of electrical energy while delivering 35 J of work each second. How much energy must be dissipated in the form of friction (heat)?
Detailed Solution for Test: First Law of Thermodynamics - Question 9

We know that power is energy per second. So by this definition, 187W means 187J/s energy is given. 
And 35J work is done.
So energy dissipated in the form of friction= 187 - 35 = 152J

Test: First Law of Thermodynamics - Question 10
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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,
*Multiple options can be correct
Test: First Law of Thermodynamics - Question 11
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Direction (Q. Nos. 11-15) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.

Q. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the following figure. Select correct statement.

[IIT JEE 2012]
Detailed Solution for Test: First Law of Thermodynamics - Question 11

T1 = T2 because process is isothermal
Work done in adiabatic process is less than in by isothermal process as area under isothermal curve is more than adiabatic curve.
In adiabatic processes, expansion is done by using internal energy. Hence it decreases while in the isothermal process, temperature remains constant. So   there is no change in internal energy.

*Multiple options can be correct
Test: First Law of Thermodynamics - Question 12
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A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375.0 mL at constant temperature of 37.0°C. As it does, it absorbs 208 J of heat. The value of q and W for the process will be (R = 8.314 J mol-1 K-1, In 7.5 =2.01)

[JEE Main 2013]
Detailed Solution for Test: First Law of Thermodynamics - Question 12

*Multiple options can be correct
Test: First Law of Thermodynamics - Question 13
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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.
Detailed Solution for Test: First Law of Thermodynamics - Question 13

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.

*Multiple options can be correct
Test: First Law of Thermodynamics - Question 14
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An ideal gas in a thermally insulated vessel at internal pressure p1 volume V1 and absolute temperature T1 expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute tem perature of the gas are p2, V2 and T2, respectively. For this expansion

[JEE Advanced 2014]
Detailed Solution for Test: First Law of Thermodynamics - Question 14

Since the vessel is thermally insulated, q = 0
Further since,Pext = 0, so w = 0,hence ∆U = 0
Since ∆T = 0, T,2 = T1 and P2V2 = P1V1
However, the process is adiabatic irreversible, so we can’t apply P2V2γ = P1V1γ

Test: First Law of Thermodynamics - Question 15
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In a certain chemical process, a lab technician supplies 254 J of heat to a system. At the same time, 73 J of work area done on the system by its surroundings. What is the increase in the internal energy (in J) of the system ?
Detailed Solution for Test: First Law of Thermodynamics - Question 15

1st law of thermodynamics states that:

δQ = ΔU + w

It is given that:

δQ = 254J

w = −73J because 73J of work is done on the gas

i.e. ΔU = 254 + 73 = 327J

Test: First Law of Thermodynamics - Question 16
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Direction (Q. Nos. 16-17) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d).

A fixed mass m of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure.

 

Q. The succeeding operations that enable this transformation of states are 

[JEE Advanced 2013] 
Detailed Solution for Test: First Law of Thermodynamics - Question 16

From K to L; volume increases(sample is expanding) t = heating
From L to M; pressure decreases(molecules are moving far) = cooling
From M to N; volume decreases = cooling
From N to K; pressure increasing = heating

Test: First Law of Thermodynamics - Question 17
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A fixed mass m of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure.

 

Q. A pair of isochoric processes among the transform ation of states is 
Detailed Solution for Test: First Law of Thermodynamics - Question 17

Isochoric process is that process where volume remains constant.
From the graph, we va say that for L→M and for N→K, volume remains constant.

Test: First Law of Thermodynamics - Question 18
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Direction (Q. Nos. 18) 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.

I. A sample of an ideal gas underwent an expansion against a constant external pressure of 1.0 bar from 1.0 m3, 10.0 bar, and 273 to 10.0 m3, 1.0 bar and 273 K. Work done by the system on the surroundings is W1 (in J).

II. The external pressure was increased to 10.0 bar and the gas sample was isothermally compressed from 10.0 m3 and 1.0 bar to 10 m3 and 10.0 bar.
Work done on the system by the surroundings is W2 (in J).

III. Total work done is W3 (in J).

IV. Least amount of work needed to restore the system to the original p-V conditions is W4 (in J) .

Match W1 W2, W3 and W4 in Column I with corresponding values in Column II.
*Answer can only contain numeric values
Test: First Law of Thermodynamics - Question 19
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Direction (Q. Nos. 19 and 20) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive).

Q. About 1 dm3 gastric juice are produced per day in the human body from other body fluids such as blood, whose pH is 7.6021 [H+] = 10-pH] at 53°C. Calculate the minimum work (in kJ) required to produce 1 dm3 gastric juice may be taken as unity.

Detailed Solution for Test: First Law of Thermodynamics - Question 19

Blood → Gastric Juice
pH = 7.6021
[H+] = 2.5×10-8 
∆G° = -RTlnK
= - RTln [1/2.5×10-8]      (since, concentration of gastric juice be taken as unity and we have blood in reactant)
= 47443.26
Now for 1 mole(22.4L) we have Energy = 47443.26
So, for 1dm3(1 L), we have
   =47443.26/22.4
 =2 kJ
And so according to me, the correct answer is 2 and not 4.

*Answer can only contain numeric values
Test: First Law of Thermodynamics - Question 20
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One mole of an ideal gas, Cp = 4 R at 300 K Is expanded adiabatically to n times the original volume. What is value of n if temperature falls by 150°?

Detailed Solution for Test: First Law of Thermodynamics - Question 20

Cp - Cv = R
4R - Cv = R
Cv = 3R
γ = Cp/Cv = 4/3
T1/T2 = (v2/v1)(4/3 - 1)
300/150 = (v2/v1)1/3
2 = (v2/v1)1/3
8 = v2/v1
v2 = 8v1
Therefore n = 8

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