Thermodynamics Test

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Thermodynamics Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

A gas for which $$\gamma =1.5$$ is suddenly compressed to $$\dfrac{1}{4}$$ th of its initial value then the ratio of the final to initial pressure is

A
1:16

B
1:8

C
1:4

D
8:1

Solution

Suddenly compressed implies adiabatic process
$$P_1V_1^{\gamma }=P_2V_2^{\gamma }$$
$$\displaystyle \dfrac {P_1}{P_2}=\left (\dfrac {V_2}{V_1}\right )^{\gamma }=\left (\dfrac {V/4}{V}\right )^{1.5}=\dfrac {1}{4^{3/2}}=\frac {1}{8}$$
$$\displaystyle \dfrac {P_2}{P_1}=\dfrac {8}{1}$$
Option D.
Question 2
1 / -0

During an adiabatic process, the pressure of a gas is proportional to the cube of its adiabatic temperature. The value of $$\dfrac {C_{p}}{C_{v}}$$ for that gas is :

A
$$ \dfrac{4}{3}$$

B
$$2$$

C
$$ \dfrac{5}{3}$$

D
$$ \dfrac{3}{2}$$

Solution

Given: $$P \ \alpha \ T^{3} \rightarrow (1)$$
in adiabatic process relation between P and T is :
$$P^{1-\gamma }T^{\gamma }= constant$$
$$P \ \alpha \ T^{\frac{\gamma }{\gamma -1}}\rightarrow (2)$$
And we know, for a gas $$\dfrac{C_{p}}{C_{v}}= \gamma $$
From equation 1 and 2:
$$\dfrac{\gamma }{\gamma -1}= 3$$
So, $$\gamma = \dfrac{3}{2}$$
So, $$\dfrac{C_{p}}{C_{v}}= \dfrac{3}{2}$$
Question 3
1 / -0

P-V plots for two gases adibatic processes are shown in the figure. Plots 1 and 2 should correspond respectively

A
He and $$O_{2}$$

B
$$O_{2}$$ and He

C
He and Ar

D
$$O_{2}$$ and $$N_{2}$$

Solution

For adiabatic processes, $$PV^{\gamma }=$$constant
When Pressure decrease, Volume has to increase to maintain the above relation.
Even as Volume increases, the increase is volume would be higher when the exponent $$\gamma $$ has a lower value than when the value of $$\gamma $$ is higher.
So, when $$\gamma $$ is higher, for monoatomic it is 1.6, the value of V is smaller compared to the value of V for which $$\gamma $$ is lower, diatomic it is 1.4
Hence, Plot 1 corresponds to lower $$\gamma $$-diatomic-$$O_2$$
Plot 2 corresponds to higher $$\gamma $$-monoatomic-$$He$$
Question 4
1 / -0

During an adiabatic change the density becomes $$\cfrac{1}{16}th$$ of the initial value, then $$\cfrac{P_{1}}{P_{2}}$$ is : $$\left ( \gamma =1.5 \right )$$

A
16

B
4

C
32

D
64

Solution

For ideal gas, $$PM=dRT$$

For adiabatic process, $$T^\gamma P^{1-\gamma} =constant$$

Thus, $$T^{1.5}_{1}P^{-.5}_{1} = T^{1.5}_{2}P^{-.5}_{2} $$

Also $$\dfrac { P_{1} }{ d_{1}T_{1} } =\dfrac { P_{2} }{ d_{2}T_{2} } $$

Given, $$d_{2}=\cfrac { d_{1} }{ 16 } $$

$$\dfrac{P_1}{P_2}=16 \dfrac{T_1}{T_2}$$....(i)

$$\dfrac{P_2}{P_1} =\dfrac{T_2^3}{T_1^3} $$...(ii)

On multiplying (i) and (ii), we get:

$$1=16\dfrac{T_2^2}{T_1^2}$$

$$\dfrac{T_2}{T_1}= \dfrac{1}{4}$$

$$\Rightarrow \dfrac{T_1}{T_2}= 4$$

Substituting $$\dfrac{T_1}{T_2}$$ in equation (i)

From above equations,

$$\cfrac { P_{1} }{ P_{2} }=16\times 4=64$$
Question 5
1 / -0

The pressure and density of a diatomic
gas $$\left ( \gamma =\dfrac{7}{5} \right )$$ changes adiabatically from (p,d)
to$$\left ( p^{1} ,d^{1}\right )$$. If $$\dfrac{d^{'}}{d}=32$$ then $$\dfrac{p^{'}}{p}$$ is

A
$$\dfrac{{1}}{128}$$

B
32

C
128

D
256

Solution

in an adiabatic process relation between P and V is :
$$PV^{\gamma }= constant$$
and $$V= \dfrac{M}{d}$$
so $$\dfrac{P}{d^{\gamma }}= constant$$
$$\dfrac{P_{1}}{d_{1}^{\gamma }}= \dfrac{P_{2}}{d_{2}^{\gamma }}$$
so $$\dfrac{P^{'}}{P}= (\dfrac{d^{'}}{d})^{\gamma }$$
$$\dfrac{P^{'}}{P}= (\dfrac{1}{32})^{\dfrac{7}{5}}$$
$$\dfrac{P^{'}}{P}= 128$$
Question 6
1 / -0

Three sample of the same gas, x, y and z, for which the ratio of specific heats is $$\gamma =3/2$$ have initially the same volume. The volume of the each sample is doubled by adiabatic process in the case of x, by isobaric process in the case of y and by isothermal process in the case of z. If the initial pressure of the sample of x, y and z are in the ratio $$2\sqrt{2}:1:2$$ then the ratio of their final pressures is

A
2 : 1 : 1

B
1 : 1 : 1

C
1 : 2 : 1

D
1 : 1 : 2

Solution

For isothermal process, $$PV=$$constant

For adiabatic process, $$PV^{\gamma}=$$constant

For isobaric process, $$P=$$ constant

Let initial pressure of y be $$P$$

Then, final pressure of x=$$2\sqrt { 2 }P \times { 2 }^{ -1.5 }$$$$=P$$

final pressure of y= initial pressure of y$$=P$$

final pressure of z=$$2P\times .5$$$$=P$$

Thus ratio of final pressures is $$1:1:1$$
Question 7
1 / -0

A heat engine operates between 2100 K and 700K. Its actual efficiency is 40%. What percentage of its maximum possible efficiency is this ?

A
33.33%

B
66.67%

C
60%

D
40%

Solution

Maximum possible efficiency of a heat engine$$(\eta )=\displaystyle 1-\dfrac {T_c}{T_h}=1-\dfrac {700}{2100}=\frac {2}{3}$$
Given that the actual efficiency is $$0.4$$
Percentage of the maximum possible efficiency $$=\displaystyle \dfrac {0.4}{\dfrac {2}{3}}\times 100=\dfrac {6}{10}\times 100=60 %$$
Question 8
1 / -0

Adiabatic modulus of elasticity of a gas is $$2.1 \times 10^{5}Nm^{-2}$$. It's isothermal modulus of elasticity is ($$ \gamma =1.4$$)

A
$$1.2\times 10^{5}Nm^{-2}$$

B
$$1.4\times 10^{5}Nm^{-2}$$

C
$$1.5\times 10^{5}Nm^{-2}$$

D
$$1.8\times 10^{5}Nm^{-2}$$

Solution

Adiabatic bulk modulus of a gas $$K_S=\gamma P$$
Isothermal bulk modulus $$K_T=P$$
$$\displaystyle \frac {K_S}{K_T}=\frac {\gamma P}{P}=\gamma =\frac {C_p}{C_v}$$

$$\displaystyle \frac {2.1 \times 10^5 }{K_T}=1.4$$
$$\displaystyle K_T=\frac {2.1 \times 10^5}{1.4}=1.5 \times 10^5$$
Option C.
Question 9
1 / -0

At $$27^{o}C$$ a gas compressed suddenly such that its pressure becomes $$\dfrac{1}{32}$$ of original pressure. Final temperature in Kelvin will be $$\left ( \gamma =5/3 \right )$$

A
100

B
75

C
75.83

D
82.63

Solution

Compressed suddenly implies adiabatic process; temperature and pressure are related as
$$P^{1-\gamma }T^{\gamma }=$$constant
$$\displaystyle \left (\dfrac {P_1}{P_2}\right )^{1-\gamma }=\left (\dfrac {T_2}{T_1}\right )^{\gamma }$$
$$\displaystyle \left (\dfrac {P}{\dfrac {P}{32}}\right )^{1-\dfrac {5}{3}}=\left (\dfrac {T_2}{300}\right )^{5/3}$$
$$\displaystyle \left (32\right )^{\dfrac {-2}{3}}=\left (\dfrac {T}{300}\right )^{\dfrac {5}{3}}$$
$$\displaystyle \dfrac {1}{32^2}=\left (\dfrac {T}{300}\right )^5$$
$$\displaystyle \dfrac {T}{300}=\left (\dfrac {1}{32^2}\right )^{\dfrac {1}{5}}=\dfrac {1}{(2^5)^{2/5}}=\dfrac {1}{4}$$
$$\displaystyle T=\dfrac {300}{4}=75K$$
Option B.
Question 10
1 / -0

During an adiabatic process, if the pressure of the ideal gas is proportional to the cube of its temperature, the ratio $$\gamma =\dfrac{C_{p}}{C_{v}}$$ is
($$C_{p}=$$ Specific heat at constant pressure ; $$C_{v}=$$Specific heat at constant volume)

A
$$\dfrac{7}{5}$$

B
$$\dfrac{4}{3}$$

C
$$\dfrac{5}{3}$$

D
$$\dfrac{3}{2}$$

Solution

Given $$P\propto T^3$$
$$PT^{-3}=$$constant
For an adiabatic process, we have
$$\displaystyle PT^{\dfrac {\gamma }{1-\gamma }}=$$constant
Comparing these two equations, we get
$$\displaystyle \dfrac {\gamma }{1-\gamma }=-3$$
$$\gamma =-3+3\gamma $$
$$2\gamma =3$$
$$\displaystyle \gamma =\frac {3}{2}$$
Option D.

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Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 31

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Thermodynamics Test - 31

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SHARING IS CARING

Question 1

1:16

1:8

1:4

8:1

Solution

Question 2

$$ \dfrac{4}{3}$$

$$2$$

$$ \dfrac{5}{3}$$

$$ \dfrac{3}{2}$$

Solution

Question 3

He and $$O_{2}$$

$$O_{2}$$ and He

He and Ar

$$O_{2}$$ and $$N_{2}$$

Solution

Question 4

16

4

32

64

Solution

Question 5

$$\dfrac{{1}}{128}$$

32

128

256

Solution

Question 6

2 : 1 : 1

1 : 1 : 1

1 : 2 : 1

1 : 1 : 2

Solution

Question 7

33.33%

66.67%

60%

40%

Solution

Question 8

$$1.2\times 10^{5}Nm^{-2}$$

$$1.4\times 10^{5}Nm^{-2}$$

$$1.5\times 10^{5}Nm^{-2}$$

$$1.8\times 10^{5}Nm^{-2}$$

Solution

Question 9

100

75

75.83

82.63

Solution

Question 10

$$\dfrac{7}{5}$$

$$\dfrac{4}{3}$$

$$\dfrac{5}{3}$$

$$\dfrac{3}{2}$$

Solution

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