Thermodynamics Test

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Thermodynamics Test - 42

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Question 1
1 / -0

In the reaction given below, at constant pressure, the work done by system on surrounding is?
$$H_2O (g) \rightarrow H_2O(l)$$

A
positive

B
negative

C
no work is done

D
positive at temperatures above critical temperature

Solution

The volume of liquid is greater than vapor.
Hence, change in volume is positive
Therefore, Work done W=$$P({ V }_{ l }-{ V }_{ g })>0$$
Question 2
1 / -0

In an adiabatic process on a gas with $$\gamma=1.4$$, the pressure is increased by 0.5%. The volume decreases by about

A
0.36%

B
0.5%

C
0.7%

D
1%

Solution

from the relation $$PV^ γ=k$$
now differentiating we get
$$P γV^{ γ-1}dV+dP.V^ γ=0$$
by making proper rearrangements we get,
$$-\dfrac{dV}{V}=\dfrac{dP}{ γP}$$
In an adiabatic process on a gas with γ=1.4, the pressure is increased by 0.5%. The volume decreases by about$$-\dfrac{dV}{V}=\dfrac{0.5}{ 1.4}*100=0.36\%$$
Question 3
1 / -0

Two different ideal diatomic gases A and B are initially in the same state. A and B are then expanded to the same final volume through adiabatic and isothermal processes respectively. If $$P_A, P_B$$ and $$T_A, T_B$$ represents the final pressure and temperature of A and B respectively then

A
$$P_A < P_B$$ and $$T_A < T_B$$

B
$$P_A > P_B$$ and $$T_A > T_B$$

C
$$P_A > P_B$$ and $$T_A < T_B$$

D
$$P_A < P_B$$ and $$T_A > T_B$$

Solution

Initial pressure $$={ P }_{ A }$$
Initial temperature$$={ T }_{ A }$$
Final pressure $$={ P }_{ B }$$
Final temperature $$={ T }_{ B }$$
$$\therefore$$ For same volume, i.e, isochoric process,
$${ P }_{ A }<{ P }_{ B }$$
$${ T }_{ A }<{ T }_{ B }$$
Question 4
1 / -0

In the following P-V diagram of an ideal gas, AB and CD are isothermal whereas BC and DA are adiabatic process. The value of $$V_B/V_C$$ is

A
$$=V_A/V_D$$

B
$$< V_A/V_D$$

C
$$> V_A/V_D$$

D
cannot say

Solution

For isothermal process,
$$PV = constant$$
$$for AB :$$
$$ \therefore { P }_{ A }{ V }_{ A }={ P }_{ B }{ V }_{ B }$$
$$\dfrac { { P }_{ A } }{ { P }_{ B } } = \dfrac { { V }_{ B } }{ { V }_{ A } } \longrightarrow (1)$$
$$for BC :$$
$${ P }_{ B }{ V }_{ B }^{ Y } = { P }_{ C }{ V }_{ C }^{ Y } \longrightarrow (2)$$
Substituting value of $$ { P }_{ B }$$ from $$ (1) \& (2)$$
$$\dfrac { { P }_{ A }{ V }_{ A } }{ { V }_{ B } } { V }_{ B }^{ Y } = { P }_{ C }{ V }_{ C }^{ Y }$$
$${ P }_{ A }{ V }_{ A }{ V }_{ B }^{ Y-1 } = { P }_{ C }{ V }_{ C }^{ Y } \longrightarrow (3)$$
$$for CD :$$
$${ P }_{ C }{ V }_{ C }= { P }_{ D }{ V }_{ D }$$
$$ \dfrac { { P }_{ C } }{ { P }_{ D } } = \dfrac { { V }_{ D } }{ { V }_{ C } } $$
$$for DA :$$
$$ { P }_{ D }{ V }_{ D }^{ Y } = { P }_{ A }{ V }_{ A }^{ Y }$$
$$ \dfrac { { V }_{ C }{ P }_{ C } }{ { V }_{ D } } { V }_{ D }^{ Y } ={ P }_{ A }{ V }_{ A }^{ Y }$$
$$ { P }_{ C }{ V }_{ C }{ V }_{ D } = { P }_{ A }{ V }_{ A }^{ Y } \longrightarrow (4)$$
$${ P }_{ A }{ V }_{ A }^{ Y } ={ P }_{ C }{ V }_{ C }{ V }_{ D }^{ Y-1 }$$
$${ P }_{ A }{ V }_{ A }{ V }_{ B }^{ Y-1 } = { P }_{ C }{ V }_{ C }^{ Y }$$
Dividing the two we get,
$$\dfrac { { P }_{ A }{ V }_{ A }^{ Y } }{ { P }_{ A }{ V }_{ A }{ V }_{ B }^{ Y-1 } } = \dfrac { { P }_{ C }{ V }_{ C }{ V }_{ D }^{ Y-1 } }{ { P }_{ C }{ V }_{ C }^{ Y } } $$
$$\dfrac { { V }_{ A }^{ Y-1 } }{ { V }_{ B }^{ Y-1 } } = \dfrac { { V }_{ D }^{ Y-1 } }{ { V }_{ C }^{ Y-1 } } $$
$$\dfrac { { V }_{ B }^{ Y-1 } }{ { V }_{ C }^{ Y-1 } } =\dfrac { { V }_{ A }^{ Y-1 } }{ { V }_{ D }^{ Y-1 } } $$
$${ (\dfrac { { V }_{ B } }{ { V }_{ C } } ) }^{ Y-1 }={ (\dfrac { { V }_{ A } }{ { V }_{ D } } ) }^{ Y-1 }$$
$$\dfrac { { V }_{ B } }{ { V }_{ C } } =\dfrac { { V }_{ A } }{ { V }_{ D } } $$
$$(A) = \dfrac { { V }_{ A } }{ { V }_{ D } }$$
ANSWER : A.
Question 5
1 / -0

Two ideal gases A & B are going through adiabatic process. Choose the correct option.

A
both A & B are monoatomic

B
both A & B are diatomic

C
B is diatomic, A is monoatomic

D
B is monoatomic, A is diatomic

Solution

For adiabatic process, $$PV^{\gamma}= K$$
$$\implies ln P= -\gamma \ln V + \ln K$$
As slope for B > slope for A, thus $$\gamma_B> \gamma_A$$
Hence B is monoatomic and A is diatomic.
Question 6
1 / -0

An ideal gas has an initial pressure of 3 pressure units and an initial volume of 4 volume units. The table gives the final pressure and volume of the gas (in those same units) in four processes. Which processes start and end on the same isotherm

A
B
C
D
P
5
4
12
6
V
7
6
1
3

A
A

B
B

C
C

D
D

Solution

for isothermal process,
$$PV=constant$$
initially $${ P }_{ 1 }{ V }_{ 1 }=12$$
so,final $${ P }{ V }=12$$
Hence, option C is correct.
Question 7
1 / -0

The adiabatic bulk modulus of hydrogen gas $$(\gamma=1.4)$$ at NTP is

A
$$1\times 10^5 N/m^2$$

B
$$1\times 10^{-5}N/m^2$$

C
$$1.4N/m^2$$

D
$$1.4\times 10^5N/m^2$$

Solution

The adiabatic bulk modulus of electricity $$yp$$.
Where y = $$\dfrac { { C }_{ P } }{ { C }_{ V } } = 1.4$$ given here,

at NTP, P =$$ { 10 }^{ 5 }N/{ m }^{ 2 }$$
$$\therefore$$ Bulk modulus of hydrogen gas = $$1.4\times { 10 }^{ 5 }N/{ m }^{ 2 }.$$
Question 8
1 / -0

Find the amount of heat released if $$1\;kg$$ steam at $$200^{\circ}C$$ is converted into $$-20^{\circ}C$$ ice.

A
$$680\;kcal$$

B
$$600\;kcal$$

C
$$780\;kcal$$

D
$$700\;kcal$$

Solution

Heat released by steam is $$=$$ heat released when it is cooled to $$100^oC+$$ latent heat $$+$$ heat released when it is cooled to $$0^oC$$ $$+$$ latent heat of fusion.
$$Q=[1\times0.5\times(200-100)]+[540\times1]+[1\times1\times(100-0)]+[80\times1]+[1\times0.5\times(0-(-20))]=780\ kcal$$
Question 9
1 / -0

If molar heat capacity of the given process (as shown in figure) is C, then

A
$$C < C_v$$

B
$$C=0$$

C
$$C > C_v$$

D
$$C=C_v$$

Solution

$$C-C_v=R$$ [Mayer's formula]
So, $$C=R+C_v$$
Therefore,$$C>C_v$$
Question 10
1 / -0

The molar heat capacity at constant pressure for Nitrogen gas at STP is nearly 3.5 R. Now, when the temperature is increased, it gradually increases and approaches 4.5 R. The most appropriate reason for this behaviour is that at high temperatures

A
nitrogen does not behave as an ideal gas

B
nitrogen molecules dissociate in atoms

C
the molecules collides more frequently

D
molecular vibration gradually become effective

Solution

The molar heat capacity at constant pressure for Nitrogen gas at STP is nearly 3.5 R. Now, when the temperature is increased, it gradually increases and approaches 4.5 R. The most appropriate reason for this behaviour is that at high temperatures molecular vibration gradually become effective.

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Thermodynamics Test - 42

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Thermodynamics Test - 42

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Question 1

positive

negative

no work is done

positive at temperatures above critical temperature

Solution

Question 2

0.36%

0.5%

0.7%

1%

Solution

Question 3

$$P_A < P_B$$ and $$T_A < T_B$$

$$P_A > P_B$$ and $$T_A > T_B$$

$$P_A > P_B$$ and $$T_A < T_B$$

$$P_A < P_B$$ and $$T_A > T_B$$

Solution

Question 4

$$=V_A/V_D$$

$$< V_A/V_D$$

$$> V_A/V_D$$

cannot say

Solution

Question 5

both A & B are monoatomic

both A & B are diatomic

B is diatomic, A is monoatomic

B is monoatomic, A is diatomic

Solution

Question 6

A

B

C

D

Solution

Question 7

$$1\times 10^5 N/m^2$$

$$1\times 10^{-5}N/m^2$$

$$1.4N/m^2$$

$$1.4\times 10^5N/m^2$$

Solution

Question 8

$$680\;kcal$$

$$600\;kcal$$

$$780\;kcal$$

$$700\;kcal$$

Solution

Question 9

$$C < C_v$$

$$C=0$$

$$C > C_v$$

$$C=C_v$$

Solution

Question 10

nitrogen does not behave as an ideal gas

nitrogen molecules dissociate in atoms

the molecules collides more frequently

molecular vibration gradually become effective

Solution

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