Thermodynamics Test

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Thermodynamics Test - 71

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Question 1
1 / -0

In adiabatic expansion of a gas

A
Its pressure increases

B
Its temperature falls

C
Its density increases

D
Its thermal energy increases

Solution

When an ideal gas is compressed adiabatically (Q=0), work is done on it and its temperature increases; in an adiabatic expansion, the gas does work and its temperature drops.
Question 2
1 / -0

For adiabatic process, wrong statement is

A
dQ = 0

B
dU = -dW

C
Q = constant

D
Entropy is not constant

Solution

For adiabatic process no heat is transferred hence $$dQ = 0, dU= – dW$$
As dQ = 0 means Q is constant
Question 3
1 / -0

$$1cm^{3}$$ of water at its boiling point absorbs 540 calories of heat to become steam with a volume of 167$$1cm^{3}$$. If the atmospheric pressure = $$1.013\times 10^{5}N/m^{2}$$ and the mechanical equivalent of heat = 4.19 J / calorie, the energy spent in this process in overcoming intermolecular forces is

A
540 cal

B
40 cal

C
500 cal

D
Zero

Solution
Question 4
1 / -0

We consider a thermodynamic system. If $$\bigtriangleup U$$ represents the increase in its internal energy and W the work done by the system. Which of the following statments is true

A
$$\bigtriangleup U= -W$$ in an adiabatic process

B
$$\bigtriangleup U= W$$ in an isothermal process

C
$$\bigtriangleup U= -W$$ in an isothermal process

D
$$\bigtriangleup U= W$$ in an adiabatic process

Solution

According to the first law of thermodynamics
$$ \Delta Q =\Delta U+ \Delta W $$
In adiabatic process $$ \Delta Q =0$$
so,$$\Delta U=-\Delta W$$
Question 5
1 / -0

A Carton's engine is made to work between $$200^{\circ}C$$ and $$0^{\circ}C$$ first and then between $$0^{\circ}$$C and $$200^{\circ}$$C. The ratio of efficiency of the engine in the two cases is

A
$$ 1.73 : 1 $$

B
$$ 1 : 1.73 $$

C
$$ 1 : 1 $$

D
$$ 1 : 2 $$

Solution

In first case $$\eta_1=1-\dfrac{T_{2}}{T_{1}}=1-\dfrac{(273+0)}{(273+200)}=\dfrac{200}{473}$$
In the second case, $$\eta_2=1-\dfrac{T_2}{T_1}=1-\dfrac{273+200}{273+0}=\dfrac{-200}{273}$$
so, $$\dfrac{\eta_1}{\eta_2}=1:1.73$$
Question 6
1 / -0

A gas is suddenly compressed to 1/4 th of it volume at normal temperature. The increase in its temperature is $$(\gamma = 1.5)$$

A
$$273$$ K

B
$$572$$ K

C
$$373$$ K

D
$$473$$ K

Solution

The gas is suddenly compressed means the process is adiabatic
Given: $$V_2=1/4V_1$$
$$ \therefore TV^{\gamma -1} = constant \Rightarrow T_{1}V_{1}^{\gamma -1} = T_{2}V_2^{\gamma -1}$$
change in temperature
$$ = T_{2}-T_{1}=2T_{1}-T_{1}=T_{1}=273 K $$
Question 7
1 / -0

A gas for which $$\gamma = 1.5$$ is suddenly compressed to $$\dfrac{1}{4}$$ th of the initial volume. Then the ratio of the final to the initial pressure is

A
$$1 : 16$$

B
$$1 : 8$$

C
$$1 : 4$$

D
$$8 : 1$$

Solution

Since, the gas has been suddenly compressed it means the process is adiabatic.
Using the relation
$$ P_{1}V_{1}^{\gamma } = P_{2}V_{2}^{\gamma } \Rightarrow \frac{P_{2}}{P_{1}}= \left [ \frac{V_{1}}{V_{2}} \right ]^{\gamma } = \left [ \frac{4}{3} \right ]^{3/4} = \frac{8}{1} $$
Question 8
1 / -0

An ideal heat engine working between temperature T and T has an efficiency $$\eta $$, the new efficiency if both the source and sink temperature are doubled, will be

A
$$\dfrac{\eta }{2}$$

B
$$\eta $$

C
$$2\eta $$

D
$$3\eta $$

Solution

In first case $$\eta _{1}=\dfrac{T_{1}-T_{2}}{T_{1}}$$
$$\eta _{1}=\dfrac{2T_{1}-2T_{2}}{2T_{1}}=\eta $$
Question 9
1 / -0

A scientist that the efficiency of his heat engine which operates at source temperature $$127^{\circ}C$$ and sink temperature $$27^{\circ}C$$ is 26% then

A
It is impossible

B
It is possible but less probable

C
It is quits probable

D
Data are incomplete

Solution

From the calculation of heat engine Efficiency
$$\eta _{max}=1-\frac{T_{2}}{T_{1}}=1=\frac{300}{400}=\frac{1}{4}=25$$
so, it iss impossible
Question 10
1 / -0

The volume of a gas reduced adiabatically to $$\dfrac{1}{4}$$ of its volume at $$27^{o} C$$, if the value of $$\gamma = 1.4$$, then new temperature will be

A
$$350 \times 4^{0.4} K$$

B
$$300 \times 4^{0.4} K$$

C
$$150 \times 4^{0.4} K$$

D
None of these

Solution

For adiabatic change $$ TV^{\gamma -1} $$ = constant
$$ \Rightarrow \frac{T_{2}}{T_{1}}= \frac{V_{1}}{V_{2}}^{\gamma -1} \Rightarrow T_{2}= \left ( \frac{V_{1}}{V_{2}} \right )^{\gamma -1}\times T_{1} $$
$$\Rightarrow T_{2} = \left ( \frac{V}{V/4} \right )^{1.4 -1}\times 300 = 300\times \left ( 4 \right )^{0.4}K $$

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Thermodynamics Test - 71

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Thermodynamics Test - 71

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SHARING IS CARING

Question 1

Its pressure increases

Its temperature falls

Its density increases

Its thermal energy increases

Solution

Question 2

dQ = 0

dU = -dW

Q = constant

Entropy is not constant

Solution

Question 3

540 cal

40 cal

500 cal

Zero

Solution

Question 4

$$\bigtriangleup U= -W$$ in an adiabatic process

$$\bigtriangleup U= W$$ in an isothermal process

$$\bigtriangleup U= -W$$ in an isothermal process

$$\bigtriangleup U= W$$ in an adiabatic process

Solution

Question 5

$$ 1.73 : 1 $$

$$ 1 : 1.73 $$

$$ 1 : 1 $$

$$ 1 : 2 $$

Solution

Question 6

$$273$$ K

$$572$$ K

$$373$$ K

$$473$$ K

Solution

Question 7

$$1 : 16$$

$$1 : 8$$

$$1 : 4$$

$$8 : 1$$

Solution

Question 8

$$\dfrac{\eta }{2}$$

$$\eta $$

$$2\eta $$

$$3\eta $$

Solution

Question 9

It is impossible

It is possible but less probable

It is quits probable

Data are incomplete

Solution

Question 10

$$350 \times 4^{0.4} K$$

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

$$150 \times 4^{0.4} K$$

None of these

Solution

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