Thermodynamics Test

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Thermodynamics Test - 47

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Question 1
1 / -0

Specific heat of a substance cannot be:

A
Finite

B
Infinite

C
Zero

D
Negative

Solution

Specific heat of a substance is always equal to or more then zero.
Therefore, it cannot be negative.
Hence, OPTION : D (negative).
Question 2
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The graph above represents the pressure and volume for a gas undergoing a cyclic process. Assume that Y and Z are on the same horizontal line, and Z and X are on the same vertical line. The part of the cyclic process from X to Y is an isothermal process.
If the pressure of the gas at X is $$4.0\ kPa$$, and the volume is $$6.0$$ cubic meters, and if the pressure at Y is $$8.0\ kPa$$, what is the volume of the gas at Y? (The graph is not necessarily drawn to scale)

A
$$12.0$$ cubic meters

B
$$16.0$$ cubic meters

C
$$3.0$$ cubic meters

D
$$4.0$$ cubic meters

E
$$2.0$$ cubic meters

Solution

Given : $$P_X =4.0$$ kPa $$V_X = 6.0$$ $$m^3$$ $$P_Y = 8.0$$ kPa
Process X to Y is an isothernal process : $$P_XV_X = P_YV_Y$$
$$\therefore$$ $$4.0 \times 6.0 = 8.0 \times V_Y$$ $$\implies V_Y = 3.0$$ $$m^3$$
Question 3
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For a process given by $$P=\dfrac { C }{ V } $$, where C is a constant, the work done by the ideal gas when the volume changes form $${ V }_{ 1 }$$ to $${ V }_{ 2 }$$ is

A
$$RT\ln\left( \dfrac { { V }_{ 2 } }{ { V }_{ 1 } } \right) $$

B
$$P({ V }_{ 2 }-{ V }_{ 1 })$$

C
$$RT\ln\left( \dfrac { { V }_{ 1 } }{ { V }_{ 2 } } \right) $$

D
$$P({ V }_{ 1}-{ V }_{ 2 })$$

Solution

$$Answer:-$$ A
See the image.
The calculated work in isothermal process is:
$$W=nRT\ln { \dfrac { { V }_{ 2 } }{ { V }_{ 1 } } } $$
Question 4
1 / -0

$$C_v$$ for helium gas $$(He)$$ is $$(in J \ mol^{-1} \ K^{-1})$$

A
2.5

B
10.5

C
1.5

D
12.5

Solution

The internal energy of 1 mole of a gas at temperature $$T$$ , having $$f$$ degrees of freedom is given by ,
$$U=N_{A}\times(1/2)fk_{B}T=(1/2)fRT$$ ......................eq1
where $$N_{A}=$$Avogadro's number
$$k_{B}=$$ Boltzmann's constant
from first law of thermodynamics ,
$$dU=dQ-dW$$ ............eq2
At constant volume , $$dV=0$$
hence $$dW=PdV=0$$
eq2 becomes , $$dU=dQ$$
but $$dQ=C_{V}dT$$ (for 1 mole of gas)
therefore $$dU=C_{V}dT$$
or $$C_{V}=dU/dT$$
putting the value of $$U$$ in this equation, we get
$$C_{V}=\frac{d(1/2)fRT}{dT}=\frac{f}{2}R$$
for Helium , $$f=3$$ as it is a monatomic gas ,
hence $$C_{V}=3R/2$$
we have $$R=8.31Jmol^{-1}K^{-1}$$
Therefore $$C_{V}=\frac{3\times8.31}{2}=12.5Jmol^{-1}K^{-1}$$
Question 5
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What is the function of refrigerants in heat pumps?

A
Refrigerants supply heat to the atmosphere

B
Refrigerants evaporate heat and cool the room

C
Refrigerant carries heat from the atmosphere to the room,which is to be heated

D
None of these

Solution

A refrigerant is a substance or mixture, usually a fluid, used in a heat pump and refrigeration cycle. In most cycles it undergoes phase transitions from a liquid to a gas and back again.A heat pump is a device that transfers heat energy from a source of heat to what is called a "heat sink". Heat pumps move thermal energy in the opposite direction of spontaneous heat transfer, by absorbing heat from a cold space and releasing it to a warmer one. A heat pump uses a small amount of external power to accomplish the work of transferring energy from the heat source to the heat sink.

the function of refrigerants in heat pumps refrigerant carries heat from the atmosphere to the room,which is to be heated
Question 6
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A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z:

A
must be in thermal equilibrium.

B
cannot be in thermal equilibrium.

C
may be in thermal equilibrium.

D
cant be in thermal equilibrium.

Solution

By zeroth law of thermodynamics, there are three parts to the system.
Let's say $$x, y, z$$

It is given that in the question is, $$A$$ system $$Y$$ is neither in thermal equilibrium with $$Y$$ nor with $$Z$$.

But, their is possibility by Zeroth law of theromodynamics that $$Y$$ and $$Z$$ may be in thermol eqyuilibrium.
$$\therefore$$ Correct option is C.
Question 7
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Which of the following $$p-V$$ diagrams best represents an isothermal process?

A

B

C

D

Solution

For isothermal process : $$pV = $$ constant
Thus $$p-V$$ graph is a hyperbola as shown in option C.
Question 8
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Which of the following is true in the case of an adiabatic process, where $$\gamma ={ C }_{ p }/{ C }_{ V }$$?

A
$${ p }^{ 1-\gamma }{ T }^{ \gamma }=$$ constant

B
$${ p }^{ \gamma }{ T }^{ 1-\gamma }=$$ constant

C
$${ p }{ T }^{ \gamma }=$$ constant

D
$${ p }^{ \gamma }{ T }=$$ constant

Solution

Adiabatic process : $$PV^{\gamma} = constant$$ .......(1)
Using ideal gas equation in (1), $$V = \dfrac{nRT}{P}$$
We get: $$P\times \bigg(\dfrac{nRT}{P}\bigg)^{\gamma} = constant$$
$$\implies$$ $$P^{1-\gamma}T^{\gamma} = constant$$
Question 9
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A system is taken from state A to state B along 2 different paths 1 and 2 . The heat absorbed and work done by the system along these 2 paths are $${Q}_{1}$$,$${Q}_{2}$$ and $${W}_{1}$$,$${W}_{2}$$

A
$${Q}_{1}$$ -$${W}_{1}$$ = $${Q}_{2}$$ - $${W}_{2}$$

B
$${Q}_{1}$$ + $${W}_{1}$$ = $${Q}_{2}$$ + $${W}_{2}$$

C
$${Q}_{1}$$ = $${Q}_{2}$$

D
$${W}_{1}$$ = $${W}_{2}$$

Solution

A system is taken from state A to state B along 2 different paths 1 and 2 . The heat absorbed and work done by the system along these 2 paths are $$Q_1,Q_2$$ and $$W_1,W_2$$
By first law of thermodynamics
$$Q_1+(−W_1)=Q_2+(−W_2)$$
Hence $$Q_1−W_1=Q_2−W_2$$ is correct.
Question 10
1 / -0

For a thermodynamics process to be reversible, the temperature difference between hot body and the working substance should be

A
zero

B
minimunm

C
maximum

D
infinity

Solution

In thermodynamics, a reversible process is a process whose direction can be "reversed" by inducing infinitesimal changes to some property of the system via its surroundings, while not increasing entropy. Throughout the entire reversible process, the system is in thermodynamic equilibrium with its surroundings, which means that the hot body and surroundings are at same temperature.