Thermodynamics Test

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Thermodynamics Test - 30

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Question 1
1 / -0

Choose the correct options for the following statements :
A) First law of thermodynamics specifies the conditions under which a body can use its heat energy to produce the work.
B) Second law of thermodynamics states that heat always flows from hot body to cold body by itself.

A
Both A and B are true

B
Both A and B are false

C
A is true but B is false

D
A is false B is true

Solution

First law of thermodynamics just states that $$dQ = dW + dU $$.

It does $$not$$ tell us whether the process is feasible or not neither does it provide any condition to under which the body uses any energy.
Hence, A is false.

Second law of thermodynamics states that heat always flows from hot body to cold body by itself and for heat to flow from a cold body to hot body, external work should be done.
Thus, B is true.
Question 2
1 / -0

On compressing a gas suddenly, its temperature

A
increases

B
decreases

C
remains constant

D
all the above

Solution

From the $$First \ Law \ of \ Thermodynamics, dQ = dU + dW$$
Sudden processes in ideal gases are adiabatic. So, dQ = 0.
So, dU = - dW
In compression of an ideal gas, dW < 0
So, dU > 0 hence dT > 0
Question 3
1 / -0

When we switch on the fan in a closed room, the temperature of the air molecules :

A
increases

B
decreases

C
remains unchanged

D
may increase or decrease depending on the speed of rotation of the fan

Solution

We are given a closed room and the fan is switched on. In this case we assume that the room is insulated and hence no heat transfer happens through the walls, floor and there are no energy interactions with the surroundings.

The energy input of the room is the energy that the fan consumes from electric power. The energy drawn by the fan motor is converted to kinetic energy of air molecules present in the room which in turn converts into thermal energy and hence, there is an increase in the temperature of the air molecules in the room.
Question 4
1 / -0

A polyatomic gas $$\left ( \gamma =\dfrac{4}{3} \right )$$ is compressed to $$\dfrac{1}{8}$$ of its volume adiabatically. If its initial pressure is P, the new pressure will be

A
$$8P$$

B
$$16P$$

C
$$6P$$

D
$$2P$$

Solution

In an adiabatic process we have $$P \times V^{\gamma} $$ constant .
or
$$P_f=P(\displaystyle\frac{V_1}{V_2})^{\dfrac{4}{3}}$$
Therefore if V become 1/8 . Substituting values in the equation will give final P to be 16P.
Question 5
1 / -0

A quantity of heat ‘Q’ is supplied to a mono-atomic ideal gas which expands at constant pressure. The fraction of heat that goes into work done by the gas is:

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

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

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

D
1

Solution

Quantity of heat supplied at constant pressure is given by $$nC_pdT$$.

Internal energy of the gas is given by $$nC_vdT$$.

The fraction of heat that goes into internal energy of the gas is $$\dfrac{C_v}{C_p}$$ ( and its value is $$\dfrac{3}{5}$$ for a monoatomic gas)

And the rest of energy is the work done by the gas $$=1 - \dfrac{3}{5} =\dfrac{2}{5}$$
Question 6
1 / -0

A given mass of a gas is compressed isothermally until its presssure is doubled. It is then allowed to expand adiabatically until its original volume is restored and its pressure is then found to be 0.75 of its initial pressure. The ratio of the specific heats of the gas is approximately :

A
1.20

B
1.41

C
1.67

D
1.83

Solution

let Initial value of volume,pressure and temperature of gas be $$V_{0}$$, $$P_{0}$$, $$T_{0}$$

In step one, gas is compressed isothermally, so temperature will be constant
After first step :
Pressure = $$2P_{0}$$; Temperature = $$T_{0}$$
from $$PV= nRT$$ Volume = $$V_{0}/2$$;

In 2nd step, gas is allowd to expand adiabatically to restore its original volume
After 2nd step:
Pressure = $$0.75P_{0}$$, Volume = $$V_{0}$$,
from the adiabatic process relation between P and V - $$PV^{\gamma }= constant$$
$$P_{1}V_{1}^{\gamma }= P_{2}V_{2}^{\gamma }$$
$$(\dfrac{V_{1}}{V_{2}})^{\gamma }= \dfrac{P_{2}}{P_{1}}$$

$$V_{1} = V_{0}/2$$, $$P_{1} =2P_{0}$$,
$$V_{2} = V_{0}$$ and $$P_{2} =0.75P_{0}$$

so $$(\dfrac{V_{0}/2}{V_{0}})^{\gamma }= \dfrac{0.75P_{0}}{2P_{0}}$$
$$(\dfrac{1}{2})^{\gamma }= \dfrac{3}{8}$$
on solving this:
$$\gamma = 1.41$$
Question 7
1 / -0

The pressure and density of a given mass of a diatomic gas ($$\gamma = \dfrac{7}{5}$$) change adiabatically from (P, d) to (P$$'$$, d$$'$$). If $$\dfrac{d'}{d}= 32$$, then $$\dfrac{P'}{P}$$ is
($$\gamma =$$ ratio of specific heats)

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

B
$$ \dfrac{1}{64}$$

C
$$64$$

D
$$128$$

Solution

In an adiabatic process, pressure and density are related as
$$Pd^{-\gamma }=$$constant
where d is density.

$$P_1d^{-\gamma }_1=P_2d^{-\gamma }_2$$
$$\displaystyle \dfrac {P}{P'}=\left (\dfrac {d'}{d}\right )^{-\gamma }=32^{-7/5}=2^{-7}$$
$$\displaystyle \dfrac {P'}{P}=2^7=128$$
Question 8
1 / -0

In an adiabatic change, the pressure P and temperature T of a monoatomic gas are related as $$P \ \alpha \ T^{c}$$ where c equals -

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

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

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

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

Solution

In adiabatic process, relation between P and T :
$$P^{1-\gamma }T^{\gamma }= constant$$

so $$P\ \alpha \ T^{\dfrac{\gamma }{\gamma -1}}$$

$$\gamma= \dfrac{5}{3}$$ for monatomic gas
and $$c= \dfrac{\gamma }{\gamma -1}$$

so $$c= \dfrac{\dfrac{5}{3}}{\dfrac{5}{3}-1}$$
$$C= \dfrac{5}{2}$$.
Question 9
1 / -0

A monatomic ideal gas initially at $$17^{o}C$$ is suddenly compressed to $$\dfrac{1}{8}$$ of its original volume. The temperature after compression is

A
$$17^{o}C$$

B
$$136^{o}C$$

C
$$887^{o}C$$

D
$$240^{o}C$$

Solution

Suddenly compressed implies no time for heat exchange with the surroundings. So, can be treated as an adiabatic process.

Relationship between volume and temperature is
$$TV^{\gamma -1}=$$constant
$$T_1V^{\gamma -1}_1=T_2V^{\gamma -1}_2$$
$$\displaystyle 290(V)^{\dfrac{5}{3}-1}=T_2\left (\dfrac {V}{8}\right )^{\dfrac{5}{3}-1}$$
$$\displaystyle T_2=290\left (\dfrac {V}{V/8}\right )^{2/3}=290(8)^{2/3}=1160 K=887 ^oC$$
Question 10
1 / -0

The tyre of a motor car contains air at 15 degree Celsius, if the temperature increases to 35 degree celsius, the approximate percentage increase in pressure is (ignore to expansion of tyre):

A
$$9$$

B
$$11$$

C
$$13$$

D
$$7$$

Solution

in adiabatic process relation between P and T:

$$P\propto T$$

$$\dfrac{\Delta P}{P}\times 100=\dfrac{\Delta T}{T}\times 100$$

$$T_{1}= 288K$$ and $$T_{2}= 308K$$

Percentage increment in pressure:

$$\dfrac{\Delta P}{P}\times 100= \dfrac{T_{2}-T_{1}}{T_{1}}\times 100$$

$$\% P= \dfrac{308-288}{308}\times 100$$

$$\% P=7$$

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Thermodynamics Test - 30

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Thermodynamics Test - 30

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SHARING IS CARING

Question 1

Both A and B are true

Both A and B are false

A is true but B is false

A is false B is true

Solution

Question 2

increases

decreases

remains constant

all the above

Solution

Question 3

increases

decreases

remains unchanged

may increase or decrease depending on the speed of rotation of the fan

Solution

Question 4

$$8P$$

$$16P$$

$$6P$$

$$2P$$

Solution

Question 5

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

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

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

1

Solution

Question 6

1.20

1.41

1.67

1.83

Solution

Question 7

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

$$ \dfrac{1}{64}$$

$$64$$

$$128$$

Solution

Question 8

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

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

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

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

Solution

Question 9

$$17^{o}C$$

$$136^{o}C$$

$$887^{o}C$$

$$240^{o}C$$

Solution

Question 10

$$9$$

$$11$$

$$13$$

$$7$$

Solution

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