Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The temperature at which the volume of ideal gas is zero is:

A
$$+274 K$$

B
$$-273^{o}C$$

C
$$273^{o} C$$

D
$$-273 K$$

Solution

From ideal gas equation, $$PV = nRT$$
This gives $$V = \dfrac{nRT}{P}$$
Now V tends to zero when T tends to zero where T is in Kelvin
So, volume becomes 0 only when temperature = $$ 0 K = -273 ^{\circ}$$C
Question 2
1 / -0

Two identical tumblers are filled fully with water at a certain temperature. One of the tumblers is warmed up and the other cooled down. But it is observed that the water overflows from both the tumblers. The temperature at which they are fully filled is

A
4 K

B
277 K

C
$$277^{0}$$C

D
$$-277^{0}$$C

Solution

Water has anomalous expansion from $$4^0$$C($$277^0$$K) to $$0^0$$C($$273^0$$K).
Water expands when heated from $$4^0$$C($$277^0$$K) onwards.
Since water overflows both the tumblers, the temp at which they are fully filled is $$277^0$$K
Question 3
1 / -0

At constant pressure, density of a gas is :

A
directly proportional to absolute temperature

B
inversely proportional to absolute temperature

C
independent of temperature

D
directly proportional to square root of absolute temperature
Solution
Hint: Density of a substance is given by $$\rho = \dfrac{m}{V}$$ , where $$m$$is the mass of the substance and $$V$$ is its volume.
Correction Option: B
Explanation for correct option:
- By ideal gas equation: $$PV = nRT$$, where $$P$$,$$V$$, $$T$$ and $$n$$ are the pressure, volume, temperature and moles of the gas and $$R$$ is the universal gas constant.
$$ \Rightarrow PV \alpha nRT$$
$$ \Rightarrow PV = \dfrac{m}{M} RT$$, where $$m$$ and $$M$$ are weight and molecular weight of the gas.
$$ \Rightarrow P = \dfrac{m}{V}\dfrac{{RT}}{M}$$
Here, $$\rho = \dfrac{m}{V}$$ and $$R' = \dfrac{R}{M}$$, where $$R'$$ is the specific gas constant for the gas.
$$ \Rightarrow P = \rho R'T$$
$$ \Rightarrow \rho = \dfrac{P}{{R'T}}$$
$$\therefore \rho \alpha \dfrac{1}{T}$$
Hence, at constant pressure, density of a gas is inversely proportional to absolute temperature.
Question 4
1 / -0

If the slope of P-T graph for a given mass of a gas increases, then the volume of the gas

A
Increases

B
Decreases

C
Does not change

D
May increase or decrease

Solution

From ideal gas equation, PV = nRT = $$\frac{w}{M}RT$$
slope of P-T curve = $$\frac{wR}{MV}$$
since V = constant and w = constant, we get that the slope is inversely proportional to molecular weight
Steeper the line, more is slope, less is volume.
So as slope increases, volume decreases
Question 5
1 / -0

An ideal gas is that which

A
Cannot be liquified

B
Can be easily liquified

C
Has strong inter molecular forces

D
Has a large size of molecules.

Solution

The concept of liquefaction of gases works on the principle of intermolecular forces of attraction between the gas molecules. Gases liquefy when these forces increase between the molecules, binding them together. But an ideal gas is one where the molecules do not influence each other, the explanation comes from Joule Thompson effect.
Question 6
1 / -0

The temperature at which a body does not radiate energy

A
$$0^{0}$$C

B
$$0\ K$$

C
$$273\ K$$

D
$$4^{0}$$ C

Solution

Energy radiated becomes $$0$$ only when absolute temperature becomes $$0$$.
Question 7
1 / -0

A box contains x molecules of a gas. How will the pressure of the gas be affected if the number of molecules is made 2x?

A
Pressure will decrease.

B
Pressure will remain unchanged.

C
Pressure will be doubled.

D
Pressure will become three times.

Solution

we know that from ideal gas equation, PV = nRT
now, n = number of moles of gas = $$\dfrac{N}{N_{A}}$$, where N = number of molecules of the gas and $$N_{A}$$ is the Avogadro's number
So, putting the value of n in ideal gas equation, we get
$$PV = \dfrac{N}{N_{A}}RT$$
this means that pressure of the gas is proportional to the number of molecules of the gas at constant temperature and volume
So, if we double the number of molecules of the gas (2x), then pressure will also double
So, C is the correct answer.
Question 8
1 / -0

In a gas equation, PV = RT, V refers to the volume of:

A
any amount of a gas

B
one gram mass of a gas

C
one gram mole of gas

D
one litre of gas

Solution

Given PV = RT
Comparing with the ideal gas equation, PV = nRT, we
find that here n = 1
So, number of moles = 1
Hence we must refer to molar volume
So, C is the correct answer.
Question 9
1 / -0

The relation between volume V, pressure P and absolute temperature T of an ideal gas is PV = xT, where x is a constant. The value of x depend upon

A
the mass of the gas molecule

B
the average kinetic energy of the gas molecules

C
P, V and T

D
the number of gas molecules in volume V.

Solution

From ideal gas equation we know that PV = nRT
We are given that PV = xT
Comparing the 2, we get that x = nR
Now, n = number of moles of the gas = $$\frac{N}{N_{A}}$$, where N is the number of molecules of the gas and $$N_{A}$$ is the avogadro's number
So, x = $$\frac{N}{N_{A}}R$$
Clearly $$N_{A}$$ and R are universal constants
So, x depends on N = number of molecules of the gas.
So, D is the correct answer.
Question 10
1 / -0

The universal gas constant has the units

A
dyne/$$^{0}C$$

B
erg/mole/k

C
erg-cm/k

D
watt/k

Solution

from ideal gas equation, PV = nRT
now, the dimensions of PV is same as dimension of energy (or dimension of work)
So, dimension of PV is ergs
and the dimension of n is mol, dimension of T is Kelvin
So, dimension of R = $$\dfrac{dimension\ of\ PV}{dimension\ of\ n \times dimension\ of\ Temperature}$$
$$ = \dfrac{erg}{mol-K}$$
so, B is the correct answer.

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Thermal Properties of Matter Test - 23

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Thermal Properties of Matter Test - 23

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Question 1

$$+274 K$$

$$-273^{o}C$$

$$273^{o} C$$

$$-273 K$$

Solution

Question 2

4 K

277 K

$$277^{0}$$C

$$-277^{0}$$C

Solution

Question 3

directly proportional to absolute temperature

inversely proportional to absolute temperature

independent of temperature

directly proportional to square root of absolute temperature

Solution

Question 4

Increases

Decreases

Does not change

May increase or decrease

Solution

Question 5

Cannot be liquified

Can be easily liquified

Has strong inter molecular forces

Has a large size of molecules.

Solution

Question 6

$$0^{0}$$C

$$0\ K$$

$$273\ K$$

$$4^{0}$$ C

Solution

Question 7

Pressure will decrease.

Pressure will remain unchanged.

Pressure will be doubled.

Pressure will become three times.

Solution

Question 8

any amount of a gas

one gram mass of a gas

one gram mole of gas

one litre of gas

Solution

Question 9

the mass of the gas molecule

the average kinetic energy of the gas molecules

P, V and T

the number of gas molecules in volume V.

Solution

Question 10

dyne/$$^{0}C$$

erg/mole/k

erg-cm/k

watt/k

Solution

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