Thermodynamics Test

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Thermodynamics Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

An ideal gas at a pressure of $$1$$ atmosphere and temperature of $$27^{o}C$$ is compressed adiabatically until its pressure becomes $$8$$ times the initial pressure, then the final temperature is
($$\gamma =3/2$$)

A
$$627^{0}\ C$$

B
$$527^{0}\ C$$

C
$$427^{0}\ C$$

D
$$327^{0}\ C$$

Solution

In an adiabatic process, pressure and temperature are related as
$$\displaystyle PT^{\dfrac {\gamma }{\gamma -1}}=$$constant
$$\displaystyle \left (\dfrac {P_1}{P_2}\right )^{1-\gamma }=\left (\dfrac {T_2}{T_1}\right )^{\gamma }$$
$$\displaystyle \left (\dfrac {1}{8}\right )^{1-1.5}=\left (\dfrac {T_2}{300}\right )^{1.5}$$
$$\displaystyle T_2=300\times {8}^{0.5/1.5}=300(2)=600K=327^o C$$
Option D.
Question 2
1 / -0

An ideal gas at $$27^{o}C$$ is compressed adiabatically to $$ \dfrac{8}{27}$$ of its original volume. The rise in temperature is (Take $$\gamma =5/3$$)

A
$$475 K$$

B
$$150K$$

C
$$275K$$

D
$$375K$$

Solution

In an adiabatic process,
$$TV^{\gamma -1}=$$constant
$$\displaystyle \dfrac {T_1}{T_2}=\left (\dfrac {V_2}{V_1}\right )^{\gamma -1}$$
$$\displaystyle \dfrac {300}{T_2}=\left (\dfrac {\dfrac {8}{27}V}{V}\right )^{\dfrac {5}{3} -1}=\left (\dfrac {8}{27}\right )^{\dfrac {2}{3}}=\frac {4}{9}$$
$$\displaystyle T_2=300\times \frac {9}{4}=675K$$
$$\therefore $$Rise in temperature$$=675-300=375K$$
Option D.
Question 3
1 / -0

A man of 60 kg gets 1000 cal of heat by eating 5 mangoes. His efficiency is 28%. To what height he can jump by using this energy

A
2m

B
20m

C
28m

D
0.2m

Solution

Energy available with the man to jump,
= $$ 0.28 \times 1000 $$ = 280 cal
If the man converts this entire energy into his potential energy,
$$ mgh = 280 $$
$$ 60 \times 9.8 \times h = 280 \times 4.2 $$
Thus, $$h = 2m$$
Question 4
1 / -0

Three samples of the same gas A, B and $$C\left ( \gamma =3/2 \right )$$ have initially equal volume. Now the volume of each sample is doubled. The process is adiabatic for A, isobaric for B and isothermal for C. If the final pressures are equal for all the three samples, the ratio of their initial pressures is:

A
$$2\sqrt{2}:2:1$$

B
$$2\sqrt{2}:1:2$$

C
$$\sqrt{2}:1:2$$

D
$$2:1:\sqrt{2}$$

Solution

For isobaric process, pressure will remain same $$P_B=P$$.
For isothermal proces,$$P_1V_1=P_2V_2\Rightarrow P_C\times V=P\times 2V$$
So $$P_C=2P$$
For adiabatic process, $$PV^{\gamma }\Rightarrow P_AV^{1.5}=P(2V)^{1.5}$$
$$P_A=2\sqrt{2}P$$
Therefore $$P_A : P_B : P_C = 2\sqrt{2} : 1 : 2 $$
Question 5
1 / -0

A refrigerator placed in a room at 300 K has inside temperature at 264K. How many calories of heat shall be delivered to the room for each 1 kcal of energy consumed by the refrigerator ideally

A
7.2 kcal

B
7.83 kcal

C
8.33 kcal

D
8.94 kcal

Solution

Since the refrigerator is reversible,
$$\dfrac {Q_2}{Q_1}=\dfrac {T_2}{T_1}=\dfrac{264}{300} $$
where, $$T_1$$ = room temperature, $$Q_1$$ =heat released
$$T_2$$ = refrigerator temperature, $$Q_2$$ = heat absorbed
From first law,
$$Q_1 -Q_2$$ =1 Kcal
Thus solving above equations we get, heat released= 8.33 Kcal
Question 6
1 / -0

For adiabatic expansion of a monoatomic perfect gas, the volume increases by 24%. The percentage of decrease in pressure is

A
24%

B
40%

C
48%

D
71%

Solution

For a monoatomic gas, $$ \gamma = 1.33 $$
Now, for an adiabatic process, $$ P {V}^{\gamma} = constant $$
Now, if V be the initial volume, new volume = 1.24 V

Substituting these values in the above expression,
if P be the initial volume, final volume comes out to be approximately, 0.6P
Thus the pressure has decreased by 40%
Question 7
1 / -0

A gas is taken through the cycle A B C A, as shown. What is the net work done by the gas ?

A
$$3000 J$$

B
$$1000 J$$

C
$$zero$$

D
$$2000 J$$

Solution

We know work done

= Area under P-V curve.

= $$\dfrac{1}{2}\ \times \left(4 \times 10^5\right) \times \left(5 \times 10^{3}\right)$$

$$=10\times 10^5 \times 10^{-3} = 1000J$$
Question 8
1 / -0

Water of mass $$m_2$$ = 1 kg is contained in a copper calorimeter of mass $$m_1$$ = 1 kg. Their common temperature t = $$10^{0}C$$. Now a piece of ice of mass $$m_3$$ = 2 kg and temperature is $$-11^{0}C$$ dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is. [specific heat of copper = 0.1 Kcal/ kg$$^{0}C$$, specific heat of water = 1 Kcal/kg$$^{0}C$$, specific heat of ice = 0.5 Kcal/kg$$^{0}C$$, latent heat of fusion of ice = 78.7 Kcal/kg]

A
$$0^{0}C$$

B
$$4^{0}C$$

C
$$- 4^{0}C$$

D
$$- 2^{0}C$$

Solution

Loss in heat from calorimeter + water as temperture changes from $$10^{0}C$$ to 0$$^{0}C$$
$$ = m_1C_110 + m_2C_210 = $$ $$1 1 10 + 1 0.1 10 = 11 kca$$l
Gain in heat of ice as its temperature changes from $$- 11^{0}C$$ to 0$$^{0}C$$$ kcal$
Hence ice and water will coexist at 0$$^{0}C$$ without any phase change
Question 9
1 / -0

If an ideal gas, at constant temperature and pressure, expands, then its

A
entropy increases and then decrease

B
internal energy increases

C
internal energy remains the same

D
internal energy decreases

Solution

Internal energy at any state is a function of the number of moles n and the temperature. If the volume expansion takes place at constant temperature and pressure we see that using the gas equation $$PV=nRT$$, we get $$V\propto n$$ and thus the number of moles of gas is increasing(maybe the gas is being introduced by an external agent), thus internal energy is increasing.
Question 10
1 / -0

Which statement is true about 2nd law of Thermodynamics?

A
Heat can be converted to into work without changes in system or surroundings

B
It is possible to construct a perpetual motion machine of second kind

C
All spontaneous process are thermodynamically irreversible

D
All spontaneous process are thermodynamically reversible

Solution

2nd law of Thermodynamics states that all spontaneous process are irreversible

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Thermodynamics Test - 32

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Thermodynamics Test - 32

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Question 1

$$627^{0}\ C$$

$$527^{0}\ C$$

$$427^{0}\ C$$

$$327^{0}\ C$$

Solution

Question 2

$$475 K$$

$$150K$$

$$275K$$

$$375K$$

Solution

Question 3

2m

20m

28m

0.2m

Solution

Question 4

$$2\sqrt{2}:2:1$$

$$2\sqrt{2}:1:2$$

$$\sqrt{2}:1:2$$

$$2:1:\sqrt{2}$$

Solution

Question 5

7.2 kcal

7.83 kcal

8.33 kcal

8.94 kcal

Solution

Question 6

24%

40%

48%

71%

Solution

Question 7

$$3000 J$$

$$1000 J$$

$$zero$$

$$2000 J$$

Solution

Question 8

$$0^{0}C$$

$$4^{0}C$$

$$- 4^{0}C$$

$$- 2^{0}C$$

Solution

Question 9

entropy increases and then decrease

internal energy increases

internal energy remains the same

internal energy decreases

Solution

Question 10

Heat can be converted to into work without changes in system or surroundings

It is possible to construct a perpetual motion machine of second kind

All spontaneous process are thermodynamically irreversible

All spontaneous process are thermodynamically reversible

Solution

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