Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

If the volume of the gas is to be increased by 4 times :

A
temperature and pressure must be doubled

B
at constant P, the temperature must be increased by 4 times

C
at constant T, the pressure must be increased by 4 times

D
it cannot be increased

Solution

$$PV=RT$$
At constant pressure temp has to be increased by 4 times to increase volume by 4 times.
Question 2
1 / -0

A gas at temperature 27$$^{0}$$C and pressure 30 atmospheres is allowed to expand to one atmospheric pressure. If the volume becomes 10 times its initial volumes, the final temperature becomes :

A
100$$^{o}$$C

B
373 K

C
373$$^{o}$$C

D
-173$$^{o}$$C

Solution

$$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \\ \dfrac { 30V }{ 300 } =\dfrac { 1\times 10V }{ { T }_{ 2 } } \\ { T }_{ 2 }=100K\\ { T }_{ 2 }=-173^0C$$
Question 3
1 / -0

When a body has the same temperature as that of its surroundings :

A
It does not radiate heat

B
It radiates same quantity of heat as it receives from the surroundings

C
It radiates less quantity of heat as it receives from the surroundings

D
It radiates more quantity of heat as it receives from the surroundings.

Solution

Since the body has the same temperature as that of surrounding, it is in thermal equilibrium with the surroundings.
So, there is no net exchange of heat.
So, amount of heat radiated = amount of heat received from the surroundings.
Question 4
1 / -0

The Universal gas constant may be expressed as :
a) 8.31 J/mole-K c) 2.00 J/mole-K
b) 8.31 cal/mole-K d) 2.00 cal/mole-K

A
a,c

B
a,d

C
b,c

D
b,d

Solution

In SI unit R= 8.31 J/mole-K
In calorie units R = 2 cal/mole-K
Question 5
1 / -0

The parameter that determine the physical state of gas are :
a) Pressure b) Volume
c) Number of moles d) Temperature

A
a & b

B
a,b & c

C
a,b & d

D
a,c & d

Solution

Surely gaseous state is determined only by pressure and volume of the substance at a particular temperature.
There P,V and T are state parameters
Question 6
1 / -0

The temperature of a gas contained in a closed vessel is increased by 2 K when the pressure is increased by 2%. The initial temperature of the gas is :

A
200K

B
100K

C
200$$^{0}$$C

D
100$$^{0}$$C

Solution

In this case, T1 = T, T2 = T +2; P1 = P P2 = 1.02P
Putting these values in P1/T1 = P2/T2 , we get, T = 100K
Question 7
1 / -0

16 gm of $$O_{2}$$ gas and x gm of $$H_{2}$$ gas occupy the same volume at the same temperature and pressure. Then x is :

A
1/2gm

B
1gm

C
8gm

D
16 gm

Solution

Equal volume of gas contains equal number of moles.
16gm of oxygen = 0.5 mole.
1mole of hydrogen =2gm.
0.5mole of hydrogen = 1gm.
Question 8
1 / -0

The mass of oxygen gas (in Kilo grams) occupying a volume of 11.2 litre at a temperature 27$$^{0}$$C and a pressure of 76cm of mercury is :
(Molecular weight of oxygen = 32)

A
0.001456

B
0.01456

C
0.1456

D
1.1456

Solution

$$PV=\dfrac { m }{ M } RT\\ 11.2\times { 10 }^{ -3 }\times 1.01\times { 10 }^{ 5 }=\dfrac { m }{ 32\times { 10 }^{ -3 } } \times 8.314\times 300\\ m=0.01456$$
Question 9
1 / -0

Select the correct formula :
(where k=Boltzmann's constant, R= gas constant, n= moles, r = density, M= molecular weight, p= pressure, T= kelvin temperature, V= volume)
a) k=RN$$_{av}$$
b)$$r=\dfrac{nM}{V}$$
c)$$\dfrac{p}{r}=\dfrac{RT}{M}$$
d) R=kN$$_{av}$$

A
a,b,c

B
a,b,d

C
b,c,d

D
a,c,d

Solution

(a) $$k= R/N_{av}$$
(b) r is density it is clearly satisfying the relation. nM gives toatal mass and when divided by volume it gives density of the substance
(c)We know that , PV=nRT
PV=wRT/M
PV/w=RT/M
P/r=RT/M
(d)$$k=R/N_{av}$$ therefore, $$R=kN_{av}$$
hence (b) (c) (d) are correct
Hence option(C)
Question 10
1 / -0

In the equation PV=constant, the numerical value of constant depends upon
a) temperature b) mass of the gas
c) system of units used d) nature of the gas

A
a & b

B
b & c

C
c & d

D
all

Solution

Since PV=nRT
Therefore value of PV depends on temperature, mass of gas , nature of gas and also system of units used

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Re-Attempt Weekly Quiz Competition

Thermal Properties of Matter Test - 11

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Thermal Properties of Matter Test - 11

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Question 1

temperature and pressure must be doubled

at constant P, the temperature must be increased by 4 times

at constant T, the pressure must be increased by 4 times

it cannot be increased

Solution

Question 2

100$$^{o}$$C

373 K

373$$^{o}$$C

-173$$^{o}$$C

Solution

Question 3

It does not radiate heat

It radiates same quantity of heat as it receives from the surroundings

It radiates less quantity of heat as it receives from the surroundings

It radiates more quantity of heat as it receives from the surroundings.

Solution

Question 4

a,c

a,d

b,c

b,d

Solution

Question 5

a & b

a,b & c

a,b & d

a,c & d

Solution

Question 6

200K

100K

200$$^{0}$$C

100$$^{0}$$C

Solution

Question 7

1/2gm

1gm

8gm

16 gm

Solution

Question 8

0.001456

0.01456

0.1456

1.1456

Solution

Question 9

a,b,c

a,b,d

b,c,d

a,c,d

Solution

Question 10

a & b

b & c

c & d

all

Solution

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