Thermodynamics Test

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Thermodynamics Test - 2

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Question 1
4 / -1

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)

A
a

B
b

C
c

D
d

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
4 / -1

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

A
mass entering = mass leaving

B
mass does not enter or leave the system

C
mass entering can be more or less than the mass leaving

D
none of the mentioned

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
4 / -1

One kg of carbon produces __________ kg of carbon dioxide.

A
  11/7

B
  4/11

C
  3/7

D
  11/3

Solution

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
Question 4
4 / -1

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

A
- 300 J

B
- 100 J

C
+100 J

D
+300 J

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
4 / -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

A
W(I) = W(II)

B
W(I) < W(II)

C
W(I) > W(II)

D
W(I) = W (II) + p_exΔV

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
4 / -1

In the following case

A
a

B
b

C
c

D
d

Solution

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
Question 7
4 / -1

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]

A
4.100 kJ mol^-1

B
3.7904 kJ mol^-1

C
37.904 k J mol^-1

D
41.00 kJ mol^-1

Solution
Question 8
4 / -1

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

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
4 / -1

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)?

A
187 J

B
35 J

C
152 J

D
222 J

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
4 / -1

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°
Question 11
4 / -1

The internal energy of compressed real gas, as compared to that of the ideal gas at the same temperature is

A
less

B
more

C
sometimes less, sometimes more

D
maximum

Solution

The internal energy of a gas is due to the energy possessed due to the motion of the molecular motion and configuration of molecules.

In case of an ideal gas, these molecular collisions and attraction are assumed to be absent as per the kinetic theory. Hence, in an ideal gas that is compressed, the gas molecules will not have potential energy but only kinetic energy. Hence, all ideal gases will have equal internal energy at same temperature.

In case of a real gas, the molecules possess both types of internal energies unlike an ideal gas. While compressing it the internal energy of the molecules is lost during the collisions within the molecules and with walls of the container. Hence compression at same temperature, internal energy(real gas) < internal energy(ideal gas)
Question 12
4 / -1

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]

A
q = + 208 J, W = - 208

B
q = - 208 J, W = - 208 J

C
q = - 208 J, W = + 208 J

D
q = + 208 J, W = + 208 J

Solution
Question 13
4 / -1

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
None of these

Solution

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.
Question 14
4 / -1

The internal energy of a perfect gas is :

A
completely kinetic

B
completely potential

C
sum of potential and kinetic energy of the molecules

D
difference of kinetic and potential energy of the molecules

Solution

In a perfect gas or an ideal gas, the atoms of molecules collide in a perfectly elastic manner. That is the intermolecular attractive forces are absent and thus the internal energy is completely kinetic.

Hence, option A is correct.
Question 15
4 / -1

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 ?

A
456 J

B
327 J

C
390 J

D
342 J

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
4 / -1

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]

A
heating, cooling, heating, cooling

B
cooling, heating, cooling, heating

C
heating, cooling, cooling, heating

D
cooling, heating, heating, cooling

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
4 / -1

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

A
K to L and L to M

B
L to M and N to K

C
L to M and M to N

D
M to N and N to K

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
4 / -1

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.

A
a

B
b

C
c

D
d
Question 19
4 / -1

Energy hidden in a definite quantity of substance is:

A
Enthalpy

B
Internal energy

C
Free energy

D
Entropy

Solution

Internal energy is the hidden energy in a definite quantity of substance
Question 20
4 / -1

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°?

A
10

B
8

C
12

D
5

Solution

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