Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 29

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Question 1
1 / -0

A closed copper vessel contains water equal to half of its volume when the temperature. Of the vessel is raised to 447$$^{0}$$c the pressure of steam in the vessel is (Treat steam as an ideal gas, R=8310 J/k / mole, density of water =1000 kg/m$$^{3}$$ molecular weight of water =18 )

A
33.24 x 10$$^{7}$$ pa

B
16.62 x 10$$^{7}$$ pa

C
10.31 x 10$$^{7}$$ pa

D
8.31x 10$$^{7}$$ pa

Solution

Let V be the volume of the copper vessel.
Then considering steam as an ideal gas,
$$PV=nRT=\dfrac{m}{M}RT$$
Mass of the water in the vessel=$$m=\rho (\dfrac{V}{2})$$
$$\implies PV=\dfrac{\rho V}{2M}RT$$

$$\implies P=\dfrac{\rho RT}{2M}=16.62\times 10^7Pa$$
Question 2
1 / -0

A sample of $$O_{2}$$ gas and a sample of hydrogen gas both have the same no. of moles, the same volume and same pressure. Assuming them to be perfect gases, the ratio of temperature of oxygen gas to the temperature of hydrogen gas is :

A
1:1

B
1:2

C
2:1

D
1:4

Solution

We know, PV=nRT
since P, V , n and R are same the ratio of temperatures is also same.
Hence the ratio is 1:1
Question 3
1 / -0

The density of a gas at N.T.P. is 1.5gm/lit. its density at a pressure of 152cm of Hg and temperature 27$$^{0}$$ C

A
$$\frac{273}{100}gm/ltr$$

B
$$\frac{150}{273}gm/ltr$$

C
$$\frac{1}{273}gm/ltr$$

D
1.5 $$gm/lit$$

Solution

Since volume is inversely proportional to density,
$$\dfrac { { P }_{ 1 } }{ { { d }_{ 1 }T }_{ 1 } } =\dfrac { { P }_{ 2 } }{ { { d }_{ 2 }T }_{ 2 } } \\ \dfrac { 760 }{ 1.5\times 273 } =\dfrac { 152\times { 10 } }{ { d }_{ 2 }\times 300 } \\ { d }_{ 2 }=\dfrac { 273 }{ 100 }gm/lit $$
Question 4
1 / -0

An electric bulb of $$250cc$$ was sealed off at a pressure $$10^{-3}$$ mm of Hg and temperature $$27^{0}$$ C. The number of molecules present in the gas is

A
$$8.02 \times 10^{15}$$

B
$$6.023 \times 10^{23}$$

C
$$8.021 \times 10^{23}$$

D
$$6 \times 10^{22}$$

Solution

$$PV=nRT\\ n=\dfrac { PV }{ RT } \\ n=\dfrac { 250\times { 10 }^{ -6 }\times 133.33\times { 10 }^{ -3 } }{ 8.314\times 300 } \\ n=1.336\times { 10 }^{ -8 }\\ N=n\times { N }_{ A }\\ N=1.336\times { 10 }^{ -8 }\times 6.022\times { 10 }^{ 23 }\\ N=8.02\times { 10 }^{ 15 }$$
Question 5
1 / -0

A uniform tube with piston in the middle and containing a gas at 0$$^{0}$$C is heated to 100$$^{0}$$C at one side . If the piston moves 5 cm, find the length of the tube containing the gas at 100$$^{0}$$C.

A
16.75 cm

B
37.5 cm

C
28.25 cm

D
12.5 cm

Solution

As $$\dfrac{PV}{T}$$ is constant for gas in both sections of the cylinder.
Let the length and area of the tube be $$L$$ and $$A$$, respectively.
For section with constant temperature,
$$P_{0}V_{0}=P_1V_1$$
$$\implies P_ (A\dfrac{L}{2})=P_1(A(\dfrac{L}{2}-5))$$
Initial temperature $$T_0 = 0^oC = 273 \textrm{K}$$
Final temperature $$T_1 = 100^oC = 100+273 = 373 \textrm{K}$$
Similarly for section whose temperature increased,
$$\dfrac{P_0V_0}{T_0}=\dfrac{P_1V_1}{T_1}$$
$$\implies \dfrac{P_0(A\dfrac{L}{2})}{273}=\dfrac{P_1(A(\dfrac{L}{2}+5))}{373}$$
$$\implies L=65cm$$
So the length of tube containing gas at $$100^{\circ}$$ is $$\dfrac{L}{2}+5=37.5cm$$
Question 6
1 / -0

A flask is filled with 13 gm of an ideal gas at 27$$^{0}$$C its temperature is raised to 52$$^{0}$$C . The mass of the gas that has to be released to maintain the temperature of the gas in the flask at 52$$^{0}$$C and the pressure remaining the same is

A
2.5 gm

B
2.0 g

C
1.5 g

D
1.0 g

Solution

Since P, V , M and R are constants,
$${ m }_{ 1 }{ T }_{ 1 }={ m }_{ 2 }{ T }_{ 2 }\\ 13\times 300={ m }_{ 2 }\times 325\\ { m }_{ 2 }=12gm$$
Hence mass to be removed = 13-12= 1gm
Question 7
1 / -0

One gram mol of helium at 27$$^{0}$$ C is mixed with three gram mols of oxygen at 127$$^{0}$$ C at constant pressure. If there is no exchange of heat with the atmosphere then the final temperature will be

A
375 K

B
175K

C
475 K

D
575K

Solution

In this case, N1 = 1, N2 = 3, T1 = 300 and T2 = 400.
Putting these values in the equation T(N1 + N2) = N1T1 + N2T2
we get final temperature T = 375K
Question 8
1 / -0

A vertical cylindrical vessel separated in two parts by a frictionless piston free to move along the length of a vessel. The length of vessel is 90 cm and the piston divides the cylinder in the ratio 2 : 1. Each of the two parts contain 0.1 mole of an ideal gas. The temperature of the gas is $$27^{o}$$C in each part. The mass of the piston is

A
41.5 kg

B
8.97 kg

C
9.57 kg

D
11.57 kg

Solution

$$PV=nRT$$
$$P = nRT/V$$
$$PA=nRT/l$$
$$PA=0.1\times 8.31\times 300/0.6$$
$$PA=41.5 N$$
Therefore option(A) is correct.
Question 9
1 / -0

During an experiment an ideal gas is found to obey an additional gas law VT =constant. The gas is initially at temperature T and pressure P. When it is heated to the temperature 2T, the resulting pressure is

A
2P

B
P/2

C
4P

D
P/4

Solution

The gas follows VT=constant. But it also follows PV=nRT.
Thus $$(\dfrac{nRT}{P})T=$$constant
$$\implies \dfrac{T^2}{P}=$$constant
Therefore $$\dfrac{T^2}{P}=\dfrac{(2T)^2}{P'}$$
$$\implies P'=4P$$
Question 10
1 / -0

A cycle tube has volume $$2000 cm^3$$. Initially the tube is filled to 3/4th of its volume by air at pressure of $$105 N/m^{2}$$. It is to be inflated to a pressure of $$6 \times 10^{5} N/m^{2}$$ under isothermal conditions. The number of strokes of pump, which gives $$500 cm^3$$ air in each stroke, to inflate the tube is

A
21

B
12

C
42

D
11

Solution

In this problem, let us conserve total number of moles before and after,

$$(6X{ 10 }^{ 5 })(2000)=(105)(1500)+(101325)(500)n$$

Where n is total number of strokes of pump.
Solving this equation given $$n\approx21$$.

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Re-Attempt Weekly Quiz Competition

Thermal Properties of Matter Test - 29

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Thermal Properties of Matter Test - 29

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SHARING IS CARING

Question 1

33.24 x 10$$^{7}$$ pa

16.62 x 10$$^{7}$$ pa

10.31 x 10$$^{7}$$ pa

8.31x 10$$^{7}$$ pa

Solution

Question 2

1:1

1:2

2:1

1:4

Solution

Question 3

$$\frac{273}{100}gm/ltr$$

$$\frac{150}{273}gm/ltr$$

$$\frac{1}{273}gm/ltr$$

1.5 $$gm/lit$$

Solution

Question 4

$$8.02 \times 10^{15}$$

$$6.023 \times 10^{23}$$

$$8.021 \times 10^{23}$$

$$6 \times 10^{22}$$

Solution

Question 5

16.75 cm

37.5 cm

28.25 cm

12.5 cm

Solution

Question 6

2.5 gm

2.0 g

1.5 g

1.0 g

Solution

Question 7

375 K

175K

475 K

575K

Solution

Question 8

41.5 kg

8.97 kg

9.57 kg

11.57 kg

Solution

Question 9

2P

P/2

4P

P/4

Solution

Question 10

21

12

42

11

Solution

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