Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

An air bubble rises from bottom to top of a liquid of density 1.5 g/cm$$^{3}$$. If its volume is doubled, the depth of liquids is

A
689 cm

B
398 cm

C
760 cm

D
152 cm

Solution

In this case, $$(P + \rho g h) V$$ $$=$$ $$2V P$$
where, $$P$$ $$=$$ atm pressure
$$\rho$$ $$=$$ density of the liquid
$$h$$ $$=$$ height of the lake
$$V$$ $$=$$ volume of the bubble
Solving we get,
$$h$$ $$=$$ $$\dfrac{P}{\rho g}$$
here, $$P$$ $$=$$ 76 X 13.6 X 9.8
$$\rho$$ $$=$$ 1.5 $$gm/cm^{3}$$ $$=$$ 1500 $$kg/m^{3}$$
Putting these values we get, $$h$$ $$=$$ 689 cm
Question 2
1 / -0

A given amount of gas is heated until both its pressure and volume are doubled. If initial temperature is 27$$^{0}$$ C, its final temperature is

A
300 K

B
600 K

C
1200 K

D
900K

Solution

from ideal gas equation, PV = nRT
So, $$T = \frac{PV}{nR}$$
Since, n is constant, and R is also constant, we can say that T is directly proportional to the product of P and V
$$T \alpha PV$$
Since both volume and pressure are doubled, absolute temperature of the gas will become 4 times
Initial temperature = 27$$^{\circ}$$C = 300K
So, final temperature = 4 x 300K = 1200 K
So, C is the correct answer.
Question 3
1 / -0

One litre of helium under a pressure of 2 atm and at 27$$^{0}$$C is heated until its pressure and volume are doubled. The final temperature attained by the gas is

A
927 K

B
927$$^{0}$$C

C
1200$$^{0}$$C

D
None

Solution

from ideal gas equation, PV = nRT
So, $$T = \frac{PV}{nR}$$
Since, n is constant, and R is also constant, we can say that T is directly proportional to the product of P and V
$$T \alpha PV$$
Since both volume and pressure are doubled, absolute temperature of the gas will become 4 times
Initial temperature = 27$$^{\circ}$$C = 300K
So, final temperature = 4 x 300K = 1200 K = 927 $$^{\circ}$$ C
So, B is the correct answer.
Question 4
1 / -0

The volume of a gas at $$27^o$$C and 2 atmospheric pressure is 2 litres. If the pressure is doubled and absolute temperature is made half, the new volume of gas is

A
50 ml

B
500 ml

C
1000 ml

D
2000 ml

Solution

from ideal gas equation, PV = nRT
So, this gives $$V = \frac{nRT}{P}$$
so, V is proportional to absolute temperature and inversely proportional to the pressure of the gas
Now temperature is made half and pressure is made 2 times
So, clearly the volume becomes one-fourth of it's original value.
So, B is the correct answer.
Question 5
1 / -0

A cylinder contains gas at a pressure of 2.5 atm. Due to leakage, the pressure falls to 2 atm, after sometime. The percentage of the gas which is leaked out is

A
40

B
15

C
20

D
25

Solution

$$\frac { { P }_{ 1 } }{ { d }_{ 1 } } =\frac { { P }_{ 2 } }{ { d }_{ 2 } } \\ \frac { 2.5 }{ { d }_{ 1 } } =\frac { 2 }{ { d }_{ 2 } } \\ { d }_{ 2 }=0.8\times { d }_{ 1 }$$
Volume escaped= $$\Delta d=\left( 1-0.8 \right) { d }_{ 1 }=0.2{ d }_{ 1 }\\ Percentage\quad decrease,\\ \frac { \Delta d }{ { d }_{ 1 } } \times 100=20$$
Question 6
1 / -0

The diagram shows the graphs of pressure vs density for a given mass of an ideal gas at two temperatures T$$_{1}$$ and T$$_{2}$$ :

A
T$$_{1}$$ > T$$_{2}$$

B
T$$_{1}$$ = T$$_{2}$$

C
T$$_{1}$$ < T$$_{2}$$

D
Any of the three are possible

Solution

$$PV= nRT$$
$$P \propto \frac{1}{V}$$ is equivalent to $$P$$ $$\propto$$ $$d$$.
The slope of P-d curve is proportional to temp
From the diagram, the slopes are equal
$$ \therefore$$ temp T is equal for both.
Question 7
1 / -0

A closed vessesl contains 8 gms of oxygen and 7gm of Nitrogen. Total pressure at a certain temperature is 10 atm. When all the oxygen is removed from the system without chage in temperature then the pressure will be

A
10 x 7/15atm

B
10 x 8/15atm

C
10 x 8/16 atm

D
10 x 8/32 atm

Solution

$$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { m }_{ 1 }{ M }_{ 2 } }{ { m }_{ 2 }{ M }_{ 1 } } \\ \\ \dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { 8\times 14 }{ 7\times 32 } \\ { P }_{ 1 }=10\times \dfrac { 8 }{ 16 } $$
Question 8
1 / -0

If $$\rho $$ is the density, m is the mass of 1 molecule and K is the Boltzman constant for a gas then the pressure of the gas is:

A
$$P=\dfrac{\rho KT}{m}$$

B
$$P=\dfrac{mKT}{\rho }$$

C
$$P=\dfrac{\rho mT}{K}$$

D
$$P=\rho KmT$$

Solution

$$PV=\dfrac { m }{ M } RT\\ P\dfrac { m }{ \rho } =nkT\\ P=\dfrac { \rho nkT }{ m } ,\quad \\ if\quad n=1,\quad P=\dfrac { \rho kT }{ m } $$
Question 9
1 / -0

The volume occupied by 8 gm of oxygen at S.T.P. is

A
11.2 lit

B
22.4 lit

C
2.8 lit

D
5.6 lit

Solution

32 gm of oxygen = 1 mole = 22.4lit (standard result according to gas governing equation)
8 gm of oxygen = 1/4 mole = 22.4/4 = 5.6lit
Question 10
1 / -0

At the top of a mountain a thermo meter read 7$$^{0}$$ C and barometer reads 70 cm of Hg. At the bottom of the mountain the barometer reads 76cm of Hg and thermometer reads 27$$^{0}$$ C. The density of air at the top of mountains is ______ times the density at the bottom.

A
0.99

B
0.9

C
0.89

D
0.95

Solution

$$\dfrac{P_1}{d_1T_1}=\dfrac{P_2}{d_2T_2}$$ (Standard equation of governance)
Dirctly putting up the value,
$$ \therefore \dfrac{70}{d_1\times ( 7+273)} = \dfrac{76}{d_2\times(27+ 273)}$$
$$\dfrac{d_1}{d_2} = \cfrac{75}{76}=0.987$$

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Re-Attempt Weekly Quiz Competition

Thermal Properties of Matter Test - 26

Result

Thermal Properties of Matter Test - 26

Score

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SHARING IS CARING

Question 1

689 cm

398 cm

760 cm

152 cm

Solution

Question 2

300 K

600 K

1200 K

900K

Solution

Question 3

927 K

927$$^{0}$$C

1200$$^{0}$$C

None

Solution

Question 4

50 ml

500 ml

1000 ml

2000 ml

Solution

Question 5

40

15

20

25

Solution

Question 6

T$$_{1}$$ > T$$_{2}$$

T$$_{1}$$ = T$$_{2}$$

T$$_{1}$$ < T$$_{2}$$

Any of the three are possible

Solution

Question 7

10 x 7/15atm

10 x 8/15atm

10 x 8/16 atm

10 x 8/32 atm

Solution

Question 8

$$P=\dfrac{\rho KT}{m}$$

$$P=\dfrac{mKT}{\rho }$$

$$P=\dfrac{\rho mT}{K}$$

$$P=\rho KmT$$

Solution

Question 9

11.2 lit

22.4 lit

2.8 lit

5.6 lit

Solution

Question 10

0.99

0.9

0.89

0.95

Solution

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