Thermodynamics Test

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Thermodynamics Test - 37

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Question 1
1 / -0

Compressed air in the tube of a wheel of a cycle at normal temperature suddenly starts coming out of the puncture. The air inside

A
starts becoming hotter

B
starts becoming colder

C
remains at the same temperature

D
may become hotter or colder depending upon the amount of water vapour present

Solution

Since tube is punctured suddenly, the process is adiabatic.

In adiabatic process : $$T{ V }^{ \gamma -1 }=constant$$
$$\dfrac { { T }_{ 2 } }{ { T }_{ 1 } } ={ (\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } ) }^{ \gamma -1 }$$
since air is expanding,
$${ V }_{ 2 }>{ V }_{ 1 }\\ \therefore { T }_{ 1 }>{ T }_{ 2 }$$
Hence, option B is correct
Question 2
1 / -0

If $$C_p$$ and $$C_v$$ are the molar specific heats of a gas at constant pressure and volume respectively then the ratio of adiabatic and isothermal modulii of elasticity will be

A
$$\displaystyle \dfrac {C_p-C_v }{ C_p } $$

B
$$\displaystyle C_pC_v$$

C
$$\displaystyle \dfrac{C_v}{C_p}$$

D
$$\displaystyle \dfrac{C_p}{C_v}$$

Solution

We know that $$\displaystyle\dfrac{C_p}{C_v}$$ is $$\gamma$$.

For isothermal process we have PV=constant
Differentiating
$$PdV+VdP=0$$
or
$$P=\displaystyle\dfrac{dP}{-dV/V}=\dfrac{Stress}{strain}=E_{iso}$$=Isothermal modulus of elasticity

For adiabatic process we have $$PV^{\gamma}=constant$$
Differentiating
$$dPV^\gamma+P\gamma V^{\gamma-1}dV=0$$
or
$$\gamma P=\displaystyle\dfrac{dP}{-dV/V}=\dfrac{stress}{strain}=E_{adia}$$=Adiabatic modulus of elasticity.

Thus $$\displaystyle\dfrac{E_{adia}}{E_{iso}}=\dfrac{\gamma P}{P}=\gamma=\dfrac{C_p}{C_v}$$
Question 3
1 / -0

Four curves A, B, C and D are drawn for given mass of gas. The curves which represent adiabatic and isothermal expansion are respectively

A
A and B

B
C and D

C
B and A

D
D and C

Solution

For an isothermal process $$PV=constant$$

Differentiating we get
$$PdV+VdP=0$$
or
$$\displaystyle\dfrac{dP}{dV}=-\dfrac{P}{V}$$
or
$$\tan\theta=\displaystyle\dfrac{dP}{dV}=-\dfrac{P}{V}$$-----------------------(i)

For a adiabatic process, we have
$$PV^{\gamma}=constant$$
Differentiating we get
$$dPV{\gamma}+P\gamma V^{\gamma-1}=0$$
or
$$\displaystyle\dfrac{dP}{dV}=-\gamma\dfrac{PV^{\gamma-1}}{V^{\gamma}}=-\gamma(\displaystyle\dfrac{P}{V})$$
Thus slope is $$\tan\theta=\displaystyle\dfrac{dP}{dV}=\gamma\dfrac{P}{V}$$................(ii)

Thus, slope of adiabatic process is more than the slope of isothermal process.
Also as $$P\propto\displaystyle\dfrac{1}{V}$$. Thus. we get the answer as A.
Question 4
1 / -0

Consider a hypothetical gas temperature of which increases to $$\sqrt{2}$$ times when compressed adiabatically to half the volume. Its equation can be written as

A
$$PV^{5/3} = constant$$

B
$$PV^{7/5} = constant$$

C
$$PV^{3/2} = constant$$

D
$$PV^{4/3} = constant$$

Solution

The relation between T & V in adiabatic process is given by

$$TV^{\gamma-1} =constant$$

Given that $$T_{2}=\sqrt2T_{1}$$
$$V_{2}=\dfrac{V_{1}}{2}$$

$$T_{1}V_{1}^{\gamma-1}=T_{2}V_{2}^{\gamma-1}$$

$$\dfrac{T_{1}}{T_{2}}=\left(\dfrac{ V_{2}}{V_{1}} \right)^{\gamma-1}$$

$$\dfrac{1}{\sqrt2}=\left(\dfrac{1}{2}\right)^{\gamma-1}$$

$$\left(\dfrac{1}{2}\right)^{\dfrac{1}{2}}=\left(\dfrac{1}{2}\right)^{\gamma-1} $$

Equating on both sides,

$$\gamma-1=\dfrac{1}{2}$$

$$\gamma=\dfrac{3}{2}$$

The relation between P & V in an adiabatic process is given by $$PV^{\gamma}$$=constant

$$\because PV^{\dfrac{3}{2}}=constant$$
Question 5
1 / -0

In free expansion of a gas the internal energy of the system

A
increases

B
decreases

C
is unchanged

D
changes

Solution

Consider two vessels placed in a system which is enclosed with a thermal insulation. One vessel contains a gas and the other is evacuated. The vessels are connected by a stopcock. When suddenly the stopcock is opened., the gas rushes into the evacuated vessel and expands freely.

The process is adiabatic as the vessels are placed in thermal insulating system $$dQ=0$$. Moreover the walls of the vessel are rigid and hence no external work is performed, thus $$dW=0$$.

Using first law we get the internal energy as unchanged.
Question 6
1 / -0

When a liquid is heated retaining its liquid state, then its molecules gain

A
Kinetic energy

B
potential energy

C
heat energy

D
both kinetic and potential energy

Solution

From first law of thermodynamics , heat supplied will result in a rise in internal energy as there is no work done .

This rise in internal energy is stored in the forsm of kinetic energy as there is no change in phase .
Question 7
1 / -0

The volume of a gas is reduced to $$\dfrac{1}{4}$$ of its initial volume adiabatically at $$27^o$$. The final temperature of the gas (if $$\gamma= 1.4$$) will be

A
$$27 \times (4)^{0.4}K$$

B
$$300 \times (1/4)^{0.4}K$$

C
$$100 \times (4)^{0.4}K$$

D
$$300 \times (4)^{0.4}K$$

Solution

The equation of state for an adiabatic process:
$$TV^{\gamma - 1} = constant$$
i.e. $$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1}$$

Given:
$$T_{1} = 300K$$, $$V_{2} = \dfrac{V_{1}}{4}$$

$$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1}$$
$$300\times V_{1}^{0.4} = T_{2}\dfrac{V_{1}^{0.4}}{4}$$
$$T_{2} = 300 \times (4)^{0.4}$$
Question 8
1 / -0

The relation between $$P$$ and $$T$$ for monoatomic gas during adiabatic process is $$P \propto T$$ $$^c$$. The value of c is

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

B
$$\dfrac{2}{5}$$

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

D
$$\dfrac{5}{2}$$

Solution

For an adiabatic process we have $$PT^{(\dfrac{\gamma}{1- \gamma})}=constant$$
or, $$P\propto \dfrac{1}{T^{(\dfrac{\gamma}{ 1- \gamma})}}$$

$$P \propto {T^{\dfrac{ -\gamma}{1- \gamma}}}$$

Now, $$ {\dfrac{ -\gamma}{1 - \gamma}} = c$$

For a monoatomic gas we have $$\gamma$$ as $$\displaystyle\dfrac{5}{3}$$.

$$ \dfrac{\dfrac{-5}{3}}{1-\dfrac{5}{3}} = c$$

Thus we get $$c=\displaystyle\dfrac{5}{2}$$
Question 9
1 / -0

What is the value of $$\dfrac{dP}{P}$$ for adiabatic expansion of the gas?

A
$$\displaystyle \gamma \dfrac{dV}{V}$$

B
$$\displaystyle -\dfrac{dV}{V}$$

C
$$\displaystyle \dfrac{dV}{V}$$

D
$$\displaystyle -\gamma \dfrac{dV}{V}$$

Solution

For a adiabatic process we have $$PV^{\gamma}=constant$$

Differentiating we get, $$dPV{\gamma}+P\gamma V^{\gamma-1}=0$$
or,$$\displaystyle\dfrac{dP}{dV}=-\gamma\frac{PV^{\gamma-1}}{V^{\gamma}}=-\gamma(\displaystyle\dfrac{P}{V})$$

or,$$\displaystyle\dfrac{dP}{P}=-\gamma\dfrac{dV}{V}$$
Question 10
1 / -0

A gas is enclosed in a vessel of volume 1000 cc at a pressure of 72.6 cm of Hg. It is being evacuated with the help of a piston pump, which expels 10% gas in each stroke. The pressure after the second stroke will be nearest to

A
60 cm

B
55 cm

C
66 cm

D
50 cm

Solution

Since 10% of gas is expelled after each stroke,
and PV = constant (assumption for the process)
The pressure also reduces by 10%.
Initial pressure is 72.6cm of Hg.
After one stroke, 90% of 72.6cm = 65.34cm of Hg
After second stroke, 90% of 65.34cm = 58.8cm of Hg
Which is closest to 60cm of Hg.
Option A is correct.

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Thermodynamics Test - 37

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Thermodynamics Test - 37

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Question 1

starts becoming hotter

starts becoming colder

remains at the same temperature

may become hotter or colder depending upon the amount of water vapour present

Solution

Question 2

$$\displaystyle \dfrac {C_p-C_v }{ C_p } $$

$$\displaystyle C_pC_v$$

$$\displaystyle \dfrac{C_v}{C_p}$$

$$\displaystyle \dfrac{C_p}{C_v}$$

Solution

Question 3

A and B

C and D

B and A

D and C

Solution

Question 4

$$PV^{5/3} = constant$$

$$PV^{7/5} = constant$$

$$PV^{3/2} = constant$$

$$PV^{4/3} = constant$$

Solution

Question 5

increases

decreases

is unchanged

changes

Solution

Question 6

Kinetic energy

potential energy

heat energy

both kinetic and potential energy

Solution

Question 7

$$27 \times (4)^{0.4}K$$

$$300 \times (1/4)^{0.4}K$$

$$100 \times (4)^{0.4}K$$

$$300 \times (4)^{0.4}K$$

Solution

Question 8

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

$$\dfrac{2}{5}$$

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

$$\dfrac{5}{2}$$

Solution

Question 9

$$\displaystyle \gamma \dfrac{dV}{V}$$

$$\displaystyle -\dfrac{dV}{V}$$

$$\displaystyle \dfrac{dV}{V}$$

$$\displaystyle -\gamma \dfrac{dV}{V}$$

Solution

Question 10

60 cm

55 cm

66 cm

50 cm

Solution

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