Thermodynamics Test

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Thermodynamics Test - 41

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Question 1
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Directions For Questions

A $$2000\ cc$$ sample of $$O_2$$ is confined to a cylinder. Initially, pressure of the gas is $$10^{5} N/m^2$$ and its temperature $$27^{o}C$$. The gas is subjected to a cyclic process. In the first step, its pressure is made twice at constant volume. In the second step, it expands to its initial pressure adiabatically and in the final step, the gas undergoes isobaric compression and finally attains its initial volume. Consider that the gas is ideal.

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Volume of the gas at the end of the second step (adiabatic expansion) will be nearly

A
$$4124\ cc$$

B
$$1876\ cc$$

C
$$2584\ cc$$

D
$$3272\ cc$$

Solution

$$\gamma = 7/5$$, $$P_1 = 2P, P_2 = P$$,
$$PV^{\gamma}$$ = constant
$$P_1V_1^{\gamma} = P_2V_2^{\gamma}$$
or, $$2P [2000]^{\gamma} = PV_2^{\gamma}$$
$$\displaystyle{2^{\gamma} \times 2000 = \frac{1}{2}}$$
$$V_2 = (2)^{7/5} \times 2000 = 3272\ cc$$
Question 2
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Directions For Questions

A well insulated box contains a partition dividing the box into two equal volumes as shown in figure. Initially the left hand side contains an ideal monoatomic gas and the other half is a vacuum. The partition is suddenly removed so that the gas now is contained throughout the entire box.

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How does the temperature of the gas change?

A
Increases

B
Decreases

C
Remains Same

D
Cannot be predicted

Solution

The first law of thermodynamics states that:
$$\delta Q = \Delta U +w = \Delta U +\int PdV $$
The box is well insulated; hence the heat supplied to the gas is zero.
The work done is against the pressure of a vacuum. i.e. P = 0 Pa.
So $$\delta Q = \Delta U +\int PdV $$
$$\implies \delta Q = 0 = \Delta U + 0 $$
$$\therefore$$ Change in internal energy is zero, hence the temperature of the body doesn't change.
Question 3
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For an adiabatic compression the quantity $$PV$$

A
increases

B
decreases

C
remains constant

D
depends on $$\gamma$$

Solution

The equation of state is given by:
$$TV^{\gamma - 1} = constant$$
$$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma -1 }$$
$$\dfrac{T_{2}}{T_{1}} = (\dfrac{V_{1}}{V_{2}})^{\gamma -1}$$
When the gas gets compressed, the temperature thus increases.
$$PV = nRT$$
$$P_{2}V_{2} - P_{1}V_{1} = nR(T_{2} - T_{1})$$
Since temperature increases, PV also increases.
Hence option A is correct.
Question 4
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An ideal gas of adiabatic exponent $$\gamma$$ is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Then the equation of the process in terms of the variables $$T$$ and $$V$$ is

A
$$TV^{\displaystyle \dfrac{\gamma\, -\, 1}{2}}\, =\, C$$

B
$$TV^{\displaystyle \dfrac{\gamma\, -\, 2}{2}}\, =\, C$$

C
$$TV^{\displaystyle \dfrac{\gamma\, -\, 1}{4}}\, =\, C$$

D
$$TV^{\displaystyle \dfrac{\gamma\, -\, 2}{4}}\, =\, C$$

Solution

We know that for an adiabatic process,work-done is give as
$${P}{{V}^{\gamma}}=C$$
now from ideal gas equation we know that $${PV}={MRT}$$----- ( i.e M,R=constant)
so $${P}=\dfrac{T}{V}$$
we know that $$\dfrac{{C}_{P}}{{C}_{V}}={\gamma}$$
so $$T{ V }^{ \dfrac { \gamma -1 }{ 2 } }=C$$
Question 5
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In an adiabatic change, the pressure $$P$$ and temperature $$T$$ of a diatomic gas are realtion $$P\, \propto\, T^{\alpha}$$ where $$\alpha$$ equals

A
$$1.67$$

B
$$0.4$$

C
$$0.6$$

D
$$3.5$$

Solution

The equation of state is given by:
$$T^{\gamma}P^{\gamma - 1} = constant$$
$$P^{\gamma - 1} = (constant)T^{-\gamma}$$

$$P = (constant)\ T^{(\dfrac{-\gamma}{\gamma -1})}$$
$$\alpha = \dfrac{-\gamma}{\gamma -1}$$
for a diatomic gas: $$\gamma = \dfrac{7}{5}$$
i.e. $$\alpha = -3.5$$
Hence option D is correct.
Question 6
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A mass of gas is first expanded isothermally and then compressed adiabatically to its original volume. What further simplest operation must be performed on the gas to restore it to its original state ?

A
An isobaric cooling to bring its temperature to initial value

B
An isochoric cooling to bring its pressure to its initial value

C
An isothermal process to take its pressure to its initial value

D
An isochoric heating to bring its temperature to initial value

Solution

the simplest way to bring the initial position gas we have to make the work done zero,so in iso-choric process we get $${W}_{net}=0$$,but option is also mention that heating to bring it's temperature at initial state but by heating its temperature will increase ,hence by cooling it's pressure will reduce hence it will came at initial position.
Question 7
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The given figure shows the variation of force applied by ideal gas on a piston which undergoes a process during which piston position changes from $$0.1$$ to $$0.4 m$$. If the internal energy of the system at the end of the process is $$2.5 J$$ higher, then the heat absorbed during the process is

A
$$15 J$$

B
$$17.5 J$$

C
$$20 J$$

D
$$22.5 J$$

Solution

Work Done on the system by the force is given by:
$$(\dfrac { 100-50 }{ 2 } +50)\times (0.2-0.1)\quad +\quad 50\times (0.4-0.2)\quad =\quad 75\times 0.1+50\times 0.2=\quad 17.5J$$

Therefore by First Law of Thermodynamics:
$$Q=\Delta U+W=\quad 2.5+17.5J\quad =\quad 20J$$
Question 8
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A cycle pump becomes hot near the nozzle after a few quick strokes even if they are smooth because

A
the volume of air decreases

B
the number of air molecules increase

C
the compression is adiabatic

D
collision between air particles increases

Solution

The compression that occurs is sudden and adiabatic.
To compensate for the compression of the gas (negative work). The internal energy must increase, hence the increase in temperature.
Option C is correct.
Question 9
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A gas obeys the relation $$\displaystyle V\propto \dfrac { 1 }{ { T }^{ 2 } } $$. The $$\dfrac{ C_p }{C_v}$$ value for the gas is

A
$$1.50$$

B
$$1.30$$

C
$$1.70$$

D
$$2$$

Solution

$$\displaystyle { T }^{ 2 }V=$$constant
or $$\displaystyle ({ TV }^{ { 1 }/{ 2 } }{ ) }^{ 2 }=$$constant
$$\displaystyle \because \quad { TV }^{ \gamma -1 }=$$constant
$$\displaystyle \therefore \quad \gamma -1=\dfrac { 1 }{ 2 } $$
or, $$\displaystyle \gamma =\dfrac { 3 }{ 2 } $$
Question 10
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The amount of energy required to change the temperature of $$1\ kg$$ of a substance by $$1^\circ C$$ is called _________.

A
Specific heat

B
Latent heat

C
Sensible heat

D
None of these

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Thermodynamics Test - 41

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Thermodynamics Test - 41

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Question 1

$$4124\ cc$$

$$1876\ cc$$

$$2584\ cc$$

$$3272\ cc$$

Solution

Question 2

Increases

Decreases

Remains Same

Cannot be predicted

Solution

Question 3

increases

decreases

remains constant

depends on $$\gamma$$

Solution

Question 4

$$TV^{\displaystyle \dfrac{\gamma\, -\, 1}{2}}\, =\, C$$

$$TV^{\displaystyle \dfrac{\gamma\, -\, 2}{2}}\, =\, C$$

$$TV^{\displaystyle \dfrac{\gamma\, -\, 1}{4}}\, =\, C$$

$$TV^{\displaystyle \dfrac{\gamma\, -\, 2}{4}}\, =\, C$$

Solution

Question 5

$$1.67$$

$$0.4$$

$$0.6$$

$$3.5$$

Solution

Question 6

An isobaric cooling to bring its temperature to initial value

An isochoric cooling to bring its pressure to its initial value

An isothermal process to take its pressure to its initial value

An isochoric heating to bring its temperature to initial value

Solution

Question 7

$$15 J$$

$$17.5 J$$

$$20 J$$

$$22.5 J$$

Solution

Question 8

the volume of air decreases

the number of air molecules increase

the compression is adiabatic

collision between air particles increases

Solution

Question 9

$$1.50$$

$$1.30$$

$$1.70$$

$$2$$

Solution

Question 10

Specific heat

Latent heat

Sensible heat

None of these

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