Thermodynamics Test

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Thermodynamics Test - 72

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Weekly Quiz Competition

Question 1
1 / -0

For an adiabatic expansion of a perfect gas, the value of $$\dfrac{\bigtriangleup P}{P}$$ is to equal

A
$$- \sqrt{\gamma } \dfrac{\bigtriangleup V}{V}$$

B
$$-\dfrac{\bigtriangleup V}{V}$$

C
$$-\gamma \dfrac{\bigtriangleup V}{V}$$

D
$$-\gamma^{2} \dfrac{\bigtriangleup V}{V}$$

Solution

According to the adiabatic expansion
$$ PV^{\gamma } $$= constant : Differentiating both sides
$$P\gamma V^{\gamma -1} dV+dP=0\Rightarrow \dfrac{dP}{P}= -\gamma \dfrac{dV}{V} $$
Question 2
1 / -0

An engine is supposed to operate between two reservoirs at temperature $$227^{o} C$$. The maximum possible efficiency of such an engine is

A
$$1/2$$

B
$$1/4$$

C
$$3/4$$

D
$$1$$

Solution

The efficiency of a engine is given by
$$ \eta =\dfrac{T_{1}-T_{2}}{T_{1}}= \dfrac{\left ( 273+727 \right )-\left ( 273+227 \right )}{273+727} $$=$$\dfrac{1}{2}$$
Question 3
1 / -0

Which of the accompanying PV, diagram best represents an isothermal process

A

B

C

D

Solution

In isothermal process $$P \alpha \dfrac{1}{V}$$
Hence graph between P and V is a hyperbola.
Question 4
1 / -0

The volume of air increases by 5% in its adiabatic expansion. The percentage decrease in its pressure will be

A
5%

B
6%

C
7%

D
8%

Solution

In an adibatic expansion,

$$PV^{\gamma}=constant$$
Differentiating the above equation,$$dP.V^{\gamma}=\gamma V^{\gamma-1}.PdV=0$$

$$\dfrac{dP}{P}=-{\gamma}\dfrac{dV}{V}$$
Air is primarily a diatomic gas, so
$\Rightarrow$$\dfrac{dP}{P}=-1.4\times 5=7%$$$
Question 5
1 / -0

P-V diagram of a diatomic gas is a straight line passing through origin . The molar heat capacity of the gas in the process will be

A
4 R

B
2.5 R

C
3 R

D
$$\dfrac{4R}{3}$$

Solution
Question 6
1 / -0

An ideal gas expands in such a manner that its pressure and volume can be related by equation $$PV^{2}$$= constant. During this process, the gas is

A
Heated

B
cooled

C
Neither heated nor cooled

D
First and then cooled

Solution

For an isothermal process, PV= constant and for the given process $$PV^2= constant$$. Therefore has is cooled because volume expands by a greater exponent than is in an isothermal porcess.
Question 7
1 / -0

Initial pressure and volume of a gas are P and V respectively. First it is expanded isothermally to volume 4V and then compressed adiabatically to volume V. The final p will be

A
1P

B
2P

C
4P

D
8P

Solution

1→2: $$PV=P'.4V⇒4V⇒P'=$$$$\dfrac{P}{4}$$
2→3: $$P'(4V)^{\gamma}=P'(V)^{\gamma}⇒P'=P'(4)^{\gamma}$$
$$P'=\dfrac{P}{4}(4)^{3/2}=\dfrac{P}{4}.8=2P$$
Question 8
1 / -0

Two samples A and B of gas initially at the same pressure and temperature are compressed from volume V to v/2 ( A isothermally and adiabatically.). The final pressure of A is

A
Greater than the final pressure of B

B
Equal to the final pressure of B

C
Less than the final pressure of B

D
Twice the final pressure of B

Solution

For isothermal process, PV= constant
$$P_1 V _1 =P_ 2 V _2 $$ ⟹$$P _2 =2P_ 1 $$ For adiabatic process, $$PV^{\gamma} = constant$$$$ P _1 V_ 1^{\gamma} =P_ 2 V_ 2 ^{\gamma} ⟹P_ 2 =P_ 1 ( V _2 V _1 ) ^{\gamma} ⟹P _2 =P _1 2^{\gamma} $$Since, $$\gamma$$ is always greater than 1, For the same change in volume, there will be a greater change in pressure for adiabatic process than that for isothemral process.
Question 9
1 / -0

In the following figure, four curves A, B, C and D are shown. The curves are

A
Isothermal for A and D while adiabatic for B and C

B
Adiabatic for A and C while isothermal for B and D

C
Isothermal for A and B while adiabatic for C and D

D
Isothermal for A and C while adiabatic for B and D

Solution

The adiabatic curve is steeper than the isothermal curve, in both the processes of expansion and compression. To reach a same height in a longer distance means lower slope of the line. Hence adiabatic curve is more steeper than isothermal curve
Question 10
1 / -0

The temperature - entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

A
$$1/3$$

B
$$2 / 3$$

C
$$1 / 2$$

D
$$1 / 4$$

Solution

$$Q_{1} = T_{0}S_{0} + \dfrac{1}{2} T_{0}S_{0} + \dfrac{3}{2}T_{0}S_{0}$$
$$Q_{2} = T_{0}S_{0}$$ and $$Q_{3} = 0$$
$$\eta = \dfrac{W}{Q_{1}} = \dfrac{Q_{1} - Q_{2}}{Q_{1}}$$
$$= 1 - \dfrac{Q_{2}}{Q_{1}} = 1 - \dfrac{2}{3} = \dfrac{1}{3}$$