Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

The temperature of the sun, if pressure is $$1.4 \times 10^9$$ atm, density is $$1.4 gcm^{-3}$$ and average molecular weight is 2, will be
$$[Given\ R = 8.4 J mol^{-1} K^{-1}]$$

A
$$1.2 \times 10^7 K$$

B
$$2.4 \times 10^7 K$$

C
$$0.4 \times 10^7 K$$

D
$$0.2 \times 10^7 K$$

Solution

$$PV = nRT n = \displaystyle \frac{m}{M} $$ and $$\rho = \displaystyle \frac{m}{V}$$
or
$$\displaystyle T = \frac{PV}{nR} = \frac{PM}{\rho R}$$
$$= \displaystyle \frac{1.4 \times 10^9 \times 1.01 \times 10^5 \times 2 \times 10^{-3}}{1.4 \times 10^3 \times 8.4}$$
$$= 2.4 \times 10^7 K$$
Question 2
1 / -0

Directions For Questions

An automobile tyre has volume $$0.015 m^3$$ on a cold day when temperature is $$5^o C$$ and atmospheric pressure is $$1.02\ atm$$. Under these conditions the gauge pressure is $$1.7\ atm\ (25\ lb/in^2)$$. After the car is driven on a highway for 30 minutes the temperature of the air in the tyres rises to $$45^oC$$ and volume to $$0.0159 m^3$$.

...view full instructions

What is new gauge pressure?

A
1.82 atm

B
1.70 atm

C
1.92 atm

D
none

Solution

$$\displaystyle \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$$
or $$\displaystyle P_2 = \frac{2.72 \times 150 \times 318}{278 \times 159}$$
$$= 2.94 atm$$
Gauge pressure $$= 2.94 - 1.02$$
$$= 1.92 atm$$
Question 3
1 / -0

A barometer tube 90 cm long contains some air above mercury. The reading is 74.8 cm when true atmospheric pressure is 76 cm and temperature is 30$$^o$$ C. If the reading is observed to be 75.4 cm on a day when temperature is 10$$^o$$C, then find the true pressure.

A
76.07 cm

B
75.6 cm

C
76.57 cm

D
77.123 cm

Solution

Let A be the area of the cross-section
$$V_1 = (90 - 74.8) A = 15. 2 A cm^3$$
$$P_1 = 76 - 74.8 = 1.2$$ cm of Hg
$$P_2 = (P - 75.4)$$ cm of Hg
$$V_2 = (90 - 75.4) A = 14.6 A cm^3$$
$$\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2 V_2}{T_2} = \frac{1.2 \times 15.2 \times 283}{303 \times 14.6} = 75.4 + 1.17$$
$$= 76.57$$ cm of Hg
Question 4
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Two containers of equal volume contain the same gas at pressures $$P_1$$ and $$P_2$$ and absolute temperatures $$T_1$$ and $$T_2$$, respectively. On joining the vessels, the gas reaches a common pressure $$P$$ and common temperature $$T$$. The ratio $$\displaystyle\frac{P}{T}$$ is equal to

A
$$\displaystyle\frac{P_1}{T_1}+\frac{P_2}{T_2}$$

B
$$\displaystyle\frac{P_1T_1+P_2T_2}{{(T_1+T_2)}^2}$$

C
$$\displaystyle\frac{P_1T_2+P_2T_1}{{(T_1+T_2)}^2}$$

D
$$\displaystyle\frac{P_1}{2T_1}+\frac{P_2}{2T_2}$$

Solution

$$P_1V=n_1RT_1$$
$$P_2V=n_2RT_2$$
where V is the volume of each vessel.
When the vessels are joined, $$P(2V)=(n_1 + n_2)RT$$
$$ \therefore \dfrac{P}{T}= \dfrac{1}{2}\dfrac{(n_1 + n_2)R}{V}=\dfrac{1}{2}(\dfrac{P_1}{T_1} + \dfrac{P_2}{T_2})=\dfrac{P_1}{2T_1} + \dfrac{P_2}{2T_2}$$
Question 5
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One mole of an ideal gas undergoes a process $$P = \displaystyle \frac{P_o}{1 + \left ( \displaystyle\frac{V}{V_o} \right )^2}$$, where $$P_o$$ and $$V_o$$ are constants. Find the temperature of the gas when $$V = V_o$$.

A
$$\displaystyle \frac{P_o V_o}{R}$$

B
$$\displaystyle \frac{2P_o V_o}{3R}$$

C
$$\displaystyle \frac{2 P_o V_o}{R}$$

D
$$\displaystyle \frac{P_o V_o}{2R}$$

Solution

Since the gas is ideal, therefore $$P_o V_o = RT$$

Using the relation given when $$V = V_o$$

$$P = \displaystyle \frac{P_o}{2}$$

Thus we get

$$ \displaystyle \frac{P_o}{2} (V_o) = RT$$

$$T = \displaystyle \frac{P_oV_o}{2R}$$.
Question 6
1 / -0

The African bombardier beetle stenaptinus insignis can emit a jet of defensive spray from the movable tip of its abdomen. The beetle's body has reservoirs of two different chemicals. When the beetle is disturbed, these chemicals are combined in a reaction chamber producing a compounds that is warmed from $$20^o$$C to $$100^o$$C by the heat of reaction. The high pressure produced allows the compound to the sprayed out at speeds 19 ms$$^{-1}$$ scaring away predators of all kinds. Assume specific heat of two chemicals and the spray to be same as that of water $$[4.19 \times 10^3 J (kgK)^{-1}]$$ and initial temperature of chemicals to be $$20^o$$C. How many times does the pressure increase?

A
1.5

B
2.3

C
1.9

D
1.28

Solution

$$\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}$$
or $$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{373}{293} = 1.28$$
Question 7
1 / -0

The correct graph between PV and P of one mol of gas at constant temperature will be

A

B

C

D

Solution

For an ideal gas $$PV=nRT$$
Thus for a constant temperature, $$PV=constant$$
Thus $$PV$$ does not change even when P changes.
Question 8
1 / -0

A gas is filled in a container at any temperature and at pressure 76 cm of Hg. If at the same temperature the mass of gas is increased by 50% then the resultant pressure will be

A
38 cm of Hg

B
76 cm of Hg

C
114 cm of Hg

D
152 cm of Hg

Solution

$$PV=nRT$$
$$PV=\dfrac{m}{M}RT$$
$$P=\dfrac{m}{M}\dfrac{RT}{V}$$ also given that P=76cm of Hg
Given that mass of the gas is increased by 50%
Now $$P_{1}=\left(\dfrac{m+\dfrac{m}{2}}{M}\right)\dfrac{RT}{V}$$
$$=\dfrac{3}{2}\dfrac{m}{M}\dfrac{RT}{V}$$
$$=\dfrac{3}{2}P$$
=$$\dfrac{3}{2}\times76$$
=$$114cm\ of\ Hg$$
Question 9
1 / -0

In the gas equation $$PV= RT$$, V is the volume of

A
1 mol of gas

B
1 g of gas

C
gas

D
1 litre of gas

Solution

For an Ideal Gas, $$PV=nRT$$
Here $$V$$ is the volume of n moles of gas.
Thus for $$PV=(1)RT$$, $$V$$ is the volume of 1 mol of gas.
Question 10
1 / -0

An enclosure of volume 3 litre contains 16g of oxygen, 7g of nitrogen and 11g of carbondioxide at 27$$^o$$C. The pressure exerted by the mixture is approximately

A
9 atmosphere

B
8.3 atmosphere

C
3 atmosphere

D
1 atmosphere

Solution

Total number of moles = number of moles of oxygen + number of moles of nitrogen + number of moles of carbon-di-oxide
$$\dfrac { 16 }{ 32 } +\dfrac { 7 }{ 14 } +\dfrac { 11 }{ 44 } =1mole$$
Molecular weight of Carbon-di-oxide = 44g
Molecular weight of Oxygen =32g
Molecular weight of nitrogen=14g
Using ideal gas equation,
$$P=\dfrac { nRT }{ V } =\dfrac { 1 \times8.3 \times300 }{ (3\times{ 10 }^{ -3 }) } =8.3atm(nearly)$$
Hence, Option B is correct.

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Thermal Properties of Matter Test - 35

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Thermal Properties of Matter Test - 35

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Question 1

$$1.2 \times 10^7 K$$

$$2.4 \times 10^7 K$$

$$0.4 \times 10^7 K$$

$$0.2 \times 10^7 K$$

Solution

Question 2

1.82 atm

1.70 atm

1.92 atm

none

Solution

Question 3

76.07 cm

75.6 cm

76.57 cm

77.123 cm

Solution

Question 4

$$\displaystyle\frac{P_1}{T_1}+\frac{P_2}{T_2}$$

$$\displaystyle\frac{P_1T_1+P_2T_2}{{(T_1+T_2)}^2}$$

$$\displaystyle\frac{P_1T_2+P_2T_1}{{(T_1+T_2)}^2}$$

$$\displaystyle\frac{P_1}{2T_1}+\frac{P_2}{2T_2}$$

Solution

Question 5

$$\displaystyle \frac{P_o V_o}{R}$$

$$\displaystyle \frac{2P_o V_o}{3R}$$

$$\displaystyle \frac{2 P_o V_o}{R}$$

$$\displaystyle \frac{P_o V_o}{2R}$$

Solution

Question 6

1.5

2.3

1.9

1.28

Solution

Question 7

Solution

Question 8

38 cm of Hg

76 cm of Hg

114 cm of Hg

152 cm of Hg

Solution

Question 9

1 mol of gas

1 g of gas

gas

1 litre of gas

Solution

Question 10

9 atmosphere

8.3 atmosphere

3 atmosphere

1 atmosphere

Solution

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