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QUESTION: 1

**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)**

** **

Solution:

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

QUESTION: 2

Which of the following is true for a steady flow system?

Solution:

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.

QUESTION: 3

One kg of carbon produces __________ kg of carbon dioxide.

Solution:

2CO + O2 ------------> 2CO_{2}

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 CO_{2}

56 kg of CO + 32 kg of O_{2} -------> 88 kg of CO_{2}

For, 1kg of CO + 32/56 kg of O_{2} ---------> 88/56 kg of CO_{2}

Therefore, 1kg of CO requires 32/56 or 4/7 kg of oxygen to produce 88/56kg or 11/7 kg of CO2

QUESTION: 4

A gas absorbs 200 J of heat and expands by 500 cm^{3} against a constant pressure of 2 x 10^{5} Nm^{-2}. Change in internal energy is

Solution:

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.

QUESTION: 5

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

Solution:

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.

QUESTION: 6

In the following case

Solution:

As q = 0, we have adiabatic process.

V_{1} = 100L and V_{2} = 800L

T_{1} = 300K and T_{2} = not given

For NH_{3}, γ = 4/3

Applying TV^{γ} = constant

(300)(100)^{4/3-1} = (T)(800)^{4/3-1}

T = 300/2 = 150K

W= nR(T_{2}-T_{1})/γ-1

= 1×8.314×150/(4/3-1)

= 3714 J

= 900 cal

QUESTION: 7

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]**

Solution:

QUESTION: 8

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

Thus, work done is

Solution:

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.

QUESTION: 9

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)?

Solution:

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

QUESTION: 10

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,

Solution:

*Multiple options can be correct

QUESTION: 11

**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]**

Solution:

T_{1} = T_{2} 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

QUESTION: 12

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]**

Solution:

*Multiple options can be correct

QUESTION: 13

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.

Solution:

Work done in isothermal process, w = nRTln(V_{2}/V_{1})

w_{600}/w_{300} = 1×R×600×ln(10)/ 1×R×300ln(10) = 2

∆U = 0 for isothermal processes.

*Multiple options can be correct

QUESTION: 14

An ideal gas in a thermally insulated vessel at internal pressure p_{1} volume V_{1} and absolute temperature T_{1} expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute tem perature of the gas are p_{2}, V_{2} and T_{2}, respectively. For this expansion

**[JEE Advanced 2014]**

Solution:

Since the vessel is thermally insulated, q = 0

Further since,P_{ext} = 0, so w = 0,hence ∆U = 0

Since ∆T = 0, T,2 = T_{1} and P_{2}V_{2} = P_{1}V_{1}

However, the process is adiabatic irreversible, so we can’t apply P_{2}V_{2}^{γ} = P_{1}V_{1}^{γ}

QUESTION: 15

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 ?

Solution:

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

QUESTION: 16

**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] **

Solution:

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

QUESTION: 17

**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 **

Solution:

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.

QUESTION: 18

**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 m^{3}, 10.0 bar, and 273 to 10.0 m^{3}, 1.0 bar and 273 K. Work done by the system on the surroundings is W_{1} (in J).

II. The external pressure was increased to 10.0 bar and the gas sample was isothermally compressed from 10.0 m^{3} and 1.0 bar to 10 m^{3} and 10.0 bar.

Work done on the system by the surroundings is W_{2} (in J).

III. Total work done is W_{3} (in J).

IV. Least amount of work needed to restore the system to the original p-V conditions is W_{4} (in J) .

Match W_{1} W_{2}, W_{3} and W_{4} in Column I with corresponding values in Column II.

Solution:

*Answer can only contain numeric values

QUESTION: 19

**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 dm ^{3} 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 dm^{3} gastric juice may be taken as unity.**

Solution:

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 1dm^{3}(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

QUESTION: 20

One mole of an ideal gas, C_{p} = 4 R at 300 K Is expanded adiabatically to n times the original volume. What is value of n if temperature falls by 150°?

Solution:

Cp - Cv = R

4R - Cv = R

Cv = 3R

γ = Cp/Cv = 4/3

T_{1}/T_{2} = (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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