Thermodynamics Test

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Thermodynamics Test - 55

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Question 1
1 / -0

Two spheres $$A$$ and $$B$$ with masses in the ratio $$2 : 3$$ and specific heat $$2 : 3$$ fall freely rest. If the rest in their temperatures on reaching the ground in the ratio $$1 : 2$$ the ratio of their heights of fall is:-

A
$$3 : 1$$

B
$$1 : 3$$

C
$$4 : 3$$

D
$$3 : 4$$

Solution

$$\textbf{Given:-}$$ Mass ratio of two sphere A and B = 2:3
Specific heat of two sphere A and B = 2:3
Temperature ratio = 1:2

$$\textbf{Solution:-}$$ Potential energy of sphere = Thermal energy of sphere

$$m_{A}gh_{A} = m_{A}S_{A}(\Delta T)_{A}..............(1)$$

$$ m_{B}gh_{B} = m_{B}S_{B}(\Delta T)_{B}..............(2)$$

Now divide equation 1 by 2, we get

$$ \dfrac{h_{A}}{h_{B}} = \left ( \dfrac{S_{A}}{S_{A}} \right )\times \dfrac{\Delta T_{A}}{\Delta T_{B}}$$

$$\dfrac{h_{A}}{h_{B}}= \dfrac{2}{3}\times \dfrac{1}{2}$$

$$\Rightarrow\ \dfrac{h_{A}}{h_{B}} =\ \frac{1}{3}$$

$$\textbf{Hence the correct option is B}$$
Question 2
1 / -0

An ideal heat engine working between temperatures $$T_1$$ and $$T_2$$ has an efficiency $$\eta $$ . The new efficiency if the temperatures of both the source and sink are doubled, will be

A
$$\frac{\eta }{2}$$

B
$$\eta $$

C
$$ 2 \eta $$

D
$$ 3 \eta $$

Solution

Efficiency of heat engine $$\eta=\dfrac{output}{input}=\dfrac{{T}_{1}-{T}_{2}}{{T}_{1}}$$-------(a)
Now new efficiency of heat engine when sin and source temperature is doubled$${\eta}_{new}=\dfrac{2{T}_{1}-2{T}_{2}}{2{T}_{1}}$$
$${\eta}_{new}=2\dfrac{{T}_{1}-{T}_{2}}{2{T}_{1}}=\eta$$ from a
Question 3
1 / -0

0.5 mole of diatomic gas at $$ 27^0C$$ is heated at constant pressure so that its volume is trebled. If $$ R=8.3 \, J\, mole^{-1}k^{-1}$$ then work done is

A
4980 J

B
2490 J

C
630 J

D
1245 J

Solution

$$\textbf{Given}:$$n = 0.5, T = $$27^o C$$, R = 8.3 J/mole/k

$$\textbf{Solution}:$$
Work done = $$p(V_f - V_i)$$
= $$p(3V_i - V_i)$$
= $$2pV_i$$ (Since PV = nRT)
Thus,
$$W = 2 \times n \times R \times T$$
=$$ 2 \times 0.5 \times 8.3 \times 300 = 2.49 KJ = 2490 J$$

$$\textbf{Hence B is the correct option}$$
Question 4
1 / -0

An ideal gas initially at pressure $$p_0$$ undergoes a free expansion (expansion against vaccum under adiabatic conditions) until its volume is $$3$$ times its initial volume. The gas is next adiabatically compressed back to its original volume. The pressure after compression is $${s^{\dfrac{2}{3}}{P_0}}$$. The pressure of the gas after the free expression is :

A
$$\dfrac{P_0}{3}$$

B
$${P_0}^{\dfrac{1}{3}}$$

C
$$P_0$$

D
$$3P_0$$

Solution

given:    from A to B process is a free expansion
A B
$$ Po , Vo , To $$ $$ V1 = 3Vo$$
$$\textbf{Solution}$$:

for process $$ A\rightarrow B$$
$$\Rightarrow $$ Q =0 $$--- \therefore $$adiabatic process
$$\Rightarrow $$ W = 0   $$ --- \therefore $$ free expansion
from 1st Law of thermodynamics
$$\Delta U = Q - W$$ = 0 (as $$Q $$and $$W$$ are 0 ).
we know that $$\Delta U = nCv \Delta T$$
$$\Delta U = nCv \Delta T$$ = 0 which means$$\Delta T =0$$
So $$T$$ comes out to be constant here
from ideal gas equation
$$PV =nRT$$ $$\therefore n,R,T$$are constant here
$$PV $$= Constant
$$Pa V a = Pb Vb$$

$$\dfrac{Po Vo}{3Vo} = Pb$$ = $$\dfrac{Po}{3}$$

THE CORRECT OPT: A
Question 5
1 / -0

An ideal gas $$A$$ and a real gas $$B$$ have their volumes increased from $$V$$ to $$2V$$ under isothermal conditions.The increase in internal energy

A
will be same in both $$A$$ and $$B$$

B
will be zero in both the gases

C
of $$B$$ will be more than that of $$A$$

D
of $$A$$ will be more than that of $$B$$

Solution

of B will be more than that of A.

There will be no change in the internal energy of the ideal gas.
But for the real gas, the increase in internal energy takes place
because of the work done against the intermolecular forces.
Question 6
1 / -0

If the door of a refrigerator is kept open, the room in which the refrigerator is kept

A
gets cooled

B
gets heated

C
neither gets cooled nor gets heated

D
gets cooled or heated depending on the initial temperature of the room

Solution

$$\textbf{Solution}:$$
1. If a refrigerator's door is kept open, then room will become hot, because then refrigerator exhaust more heat into the room than earlier.
2. In this way, temperature of the room increases and room becomes hot.
3. No refrigerator is efficient. Thus it exhaust more heat into the room than it extract from it.

$$\textbf{Hence B is the correct option}$$
Question 7
1 / -0

The molar heat capacity of a certain substance varies with temperature according to the given equation $$C = 27.2 + (4 \times 10^{-3})T$$. The heat necessary to change the temperature of $$2 \ mol$$ of the substance from $$300 \ K$$ to $$700 \ K$$ is:

A
$$3.46 \times 10^{4} J$$

B
$$2.33 \times 10^{3} J$$

C
$$3.46 \times 10^{3} J$$

D
$$2.33 \times 10^{4} J$$

Solution
Question 8
1 / -0

When a diatomic gas expands at constant pressure, the percentage of heat supplied that increases temperature of the gas and in doing external work in expansion at constant pressure is:

A
60%

B
40%

C
28.57 %

D
71.42 %

Solution
Question 9
1 / -0

The
molar heat capacity in a process of a diatomic gas if it does a work of $${Q
\over 4}$$ when a heat of $$Q$$ is supplied to it is

A
$${2 \over 5}R$$

B
$${5 \over 2}R$$

C
$${{10} \over 3}R$$

D
$${6 \over 7}R$$

Solution

Given: Diatomic gas doing work of $$W=\dfrac{Q}{4}$$ when $${\triangle Q=Q} $$ heat is supplied.
Solution:
From first law of thermodynamics, we know that $$\triangle Q=\triangle U+W$$
Putting the given values in above equation,we get,
$$\triangle U=\dfrac{3Q}{4}$$
We know that $${\triangle U= \dfrac{nR\triangle T}{\gamma-1}}$$ $$\rightarrow$$ (1)
The value of $$\gamma$$ for a diatomic gas is$$\dfrac{7}{5}$$. Putting in equation (1), we get,
$${n\triangle T=\dfrac{3Q}{10R}}$$ $$\rightarrow$$ (2)
Also we know that $$Q=nC\triangle T$$ $$\rightarrow$$ (3)
where $$ C $$ is molar specific heat
Putting the value of $$n\triangle T$$ from equation 2 in equation 3 we get,
$$\dfrac{3CQ}{10R}=Q$$
$$\therefore C=\dfrac{10R}{3}$$
Hence the correct option is C.
Question 10
1 / -0

A 1 h.p. electric motor operates a pump continuously. The work performed by the motor in one day is

A
18 kWh

B
36 kWh

C
54 kWh

D
72 kWh

Solution

Solution: Mathematically, $$1 h.p. = 746W$$
Also, 1 kW = $$1000W$$
So, $$1 h.p.= 0.746kW$$
Work done by the motor in one day or 24 hrs is given by,
$$E= P\times t$$
where, E denotes electrical energy,
P denotes the power in kilowatt and
t denotes the time in hours
Substituting the values in the above equation we get,
$$E=0.746kW\times 24$$
$$E=18kWh$$
Hence, the correct option is (A).

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Thermodynamics Test - 55

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Thermodynamics Test - 55

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Question 1

$$3 : 1$$

$$1 : 3$$

$$4 : 3$$

$$3 : 4$$

Solution

Question 2

$$\frac{\eta }{2}$$

$$\eta $$

$$ 2 \eta $$

$$ 3 \eta $$

Solution

Question 3

4980 J

2490 J

630 J

1245 J

Solution

Question 4

$$\dfrac{P_0}{3}$$

$${P_0}^{\dfrac{1}{3}}$$

$$P_0$$

$$3P_0$$

Solution

Question 5

will be same in both $$A$$ and $$B$$

will be zero in both the gases

of $$B$$ will be more than that of $$A$$

of $$A$$ will be more than that of $$B$$

Solution

Question 6

gets cooled

gets heated

neither gets cooled nor gets heated

gets cooled or heated depending on the initial temperature of the room

Solution

Question 7

$$3.46 \times 10^{4} J$$

$$2.33 \times 10^{3} J$$

$$3.46 \times 10^{3} J$$

$$2.33 \times 10^{4} J$$

Solution

Question 8

60%

40%

28.57 %

71.42 %

Solution

Question 9

$${2 \over 5}R$$

$${5 \over 2}R$$

$${{10} \over 3}R$$

$${6 \over 7}R$$

Solution

Question 10

18 kWh

36 kWh

54 kWh

72 kWh

Solution

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