Thermodynamics Test

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Thermodynamics Test - 16

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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)

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
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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
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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
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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

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
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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

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
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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
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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]

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
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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

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
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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)?

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
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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,

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°