Thermodynamics Test

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Thermodynamics Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

Transit of work takes place when any property other than $$\underline{\hspace{0.5in}}$$ differs

A
Temperature

B
Pressure

C
Volume

D
All the above

Solution

Zeroth law states that temperature difference leads to transit of heat energy.
Hence temperature is the required property.
Question 2
1 / -0

Heat and work are equivalent. This means,

A
When we supply heat to a body we do work on it.

B
When we do work on a body we supply heat to it.

C
The temperature of a body can be increased by doing work on it.

D
A body kept at rest may be set into motion along a line by supplying heat to it.
Question 3
1 / -0

The ratio of specific heat capacity to molar heat capacity of a body

A
is a universal constant

B
depends on the mass of the body

C
depends on the molecular weight of the body

D
is dimensionless

Solution

Specific heat capacity(C) is the amount of heat required per unit mass to raise the temperature by 1K.
Molar heat capacity(s) is the amount of heat required per mole to raise the temperature by 1K.
Thus $$\dfrac{C}{s}=\dfrac{moles}{mass}=\dfrac{1}{M_o}$$
Question 4
1 / -0

The internal energy of 1 mol of an ideal gas depends on

A
only volume

B
only temperature

C
only pressure

D
temperature and pressure

Solution

Internal energy is $$mc\Delta T$$, so it depends on temperature.
Question 5
1 / -0

That 'Entropy of a system increases in all spontaneous processes' is known as

A
1 st law of Thermodynamics

B
2nd law of Thermodynamics

C
Zeroth law of Thermodynamics

D
3rd law of Thermodynamics

Solution

The second law of thermodynamics states that the entropy of an isolated system never decreases, because isolated systems spontaneously evolve toward thermodynamic equilibriumthe state of maximum entropy
Question 6
1 / -0

The molar heat capacity of an ideal gas

A
cannot be negative

B
must be equal to either $$C_v \ or \ C_p$$

C
must lie in the range $$C_c \leq C \leq C_p$$

D
many have any value between $$- \infty $$ and $$+\infty$$
Solution
Hint: The quantity of heat energy required to increase the temperature of 1 mole of a substance is known as molar heat capacity or molar specific heat capacity.
Explanation:
- The formula of change in heat is given by: $$dQ = n C dT$$
where $$C$$ is molar heat capacity, $$n$$ is the number of moles. For various processes, values of $$dQ$$ and $$dT$$ will be different.
For the isothermal process, $$dT$$ is $$0$$. So, for non-zero value of $$dQ, C$$ will be infinite.
For the adiabatic process, change in heat $$dQ = 0$$. For that $$C$$ will be zero.
For other processes like at constant volume and at constant pressure $$C$$ will be and respectively. Which will depend on the degree of freedom that is f. So, molar heat capacity also depends on the nature of the gas and its atomicity.

So, molar heat capacity has any value between $$- \infty $$ and $$+\infty$$

And, if from saturated vapors when a certain quantity of heat is removed, at that time due to an increase in temperature specific heat will have a negative value.

Answer:

Hence, option D is the correct answer.
Question 7
1 / -0

A system is given $$400\ cal$$ of heat and $$1000\ J$$ of work is done by the system, then the change in internal energy of the system will be

A
$$-860\ J$$

B
$$680\ erg$$

C
$$680\ J$$

D
$$860\ J$$

Solution

$$1\ cal=4.184\ J, thus\ 400\ cal = 1673.6\ J$$.
The change in internal energy is $$\Delta U=\Delta Q- \Delta W$$
or $$\Delta U=1673.6-1000=673.6\approx 680\ J$$
Question 8
1 / -0

In a certain process $$500\ cal$$ of heat is given to a system and the system does $$100\ J$$ of work. The increase in internal energy of the system is

A
$$40\ cal$$

B
$$82\ cal$$

C
$$1993\ J$$

D
$$2193\ J$$

Solution

$$1\ cal=4.184\ J, thus\ 500\ cal = 2092 \ J$$.
The change in internal energy is $$\Delta U=\Delta Q- \Delta W$$
or $$\Delta U=2093-100=1993 \ J$$
Question 9
1 / -0

Which of the following is the best container for gas during adiabatic process?

A
wood vessel

B
thermos flask

C
copper vessel

D
glass vessel

Solution

A thermos flask does not allow for the heat transfer to take place through its walls.
Question 10
1 / -0

For an adiabatic expansion of perfect gas, the value of $$\displaystyle \left ( \frac{\Delta P}{P} \right )$$ is equal to

A
$$-y^{1/2} \displaystyle \left ( \frac{\Delta V}{V} \right )$$

B
$$- \displaystyle \left ( \frac{\Delta V}{V}\right )$$

C
$$- \gamma \displaystyle \left ( \frac{\Delta V}{V} \right )$$

D
$$\gamma^2 \displaystyle \left ( \frac{\Delta V}{V} \right )$$

Solution

In an adiabatic expansion of gas,
$$P{ V }^{ \gamma }=constant$$
Taking logarithm on both sides
$$ln(P)+\gamma ln(V)=ln(C)$$

On differentiating:
$$\dfrac { \Delta P }{ P } +\gamma \dfrac { \Delta V }{ V } =0\\ \Rightarrow \dfrac { \Delta P }{ P }=-\gamma \dfrac { \Delta V }{ V } $$
Hence, option C is correct.

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Thermodynamics Test - 16

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Thermodynamics Test - 16

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Question 1

Temperature

Pressure

Volume

All the above

Solution

Question 2

When we supply heat to a body we do work on it.

When we do work on a body we supply heat to it.

The temperature of a body can be increased by doing work on it.

A body kept at rest may be set into motion along a line by supplying heat to it.

Question 3

is a universal constant

depends on the mass of the body

depends on the molecular weight of the body

is dimensionless

Solution

Question 4

only volume

only temperature

only pressure

temperature and pressure

Solution

Question 5

1 st law of Thermodynamics

2nd law of Thermodynamics

Zeroth law of Thermodynamics

3rd law of Thermodynamics

Solution

Question 6

cannot be negative

must be equal to either $$C_v \ or \ C_p$$

must lie in the range $$C_c \leq C \leq C_p$$

many have any value between $$- \infty $$ and $$+\infty$$

Solution

Question 7

$$-860\ J$$

$$680\ erg$$

$$680\ J$$

$$860\ J$$

Solution

Question 8

$$40\ cal$$

$$82\ cal$$

$$1993\ J$$

$$2193\ J$$

Solution

Question 9

wood vessel

thermos flask

copper vessel

glass vessel

Solution

Question 10

$$-y^{1/2} \displaystyle \left ( \frac{\Delta V}{V} \right )$$

$$- \displaystyle \left ( \frac{\Delta V}{V}\right )$$

$$- \gamma \displaystyle \left ( \frac{\Delta V}{V} \right )$$

$$\gamma^2 \displaystyle \left ( \frac{\Delta V}{V} \right )$$

Solution

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