Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 24

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Question 1
1 / -0

The density of an ideal gas

A
is directly proportional to its pressure and absolute temperature

B
is directly proportional to its pressure and inversely proportional to its absolute temperature

C
is inversely proportional to its pressure and directly proportional to its absolute temperature

D
is inversely proportional to both its pressure and absolute temperature of the gas

Solution

Let us derive the expression for the density of an ideal gas
density = $$\frac{Mass}{Volume} = \frac{w}{V}$$
Now from ideal gas equation, PV = nRT
and n = $$\frac{w}{M}$$ where w is the mass of the gas and M is the molar mass of the gas
Putting this in ideal gas equation, we get
$$PV = \frac{w}{M}RT$$
now substitute the value of V in the density expression to get
$$d = \frac{PM}{RT}$$
so, density of an ideal gas is proportional to the pressure and inversely proportional to the absolute temperature
So, B is the correct answer.
Question 2
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If pressure and temperature of an ideal gas are doubled and volume is halved, the number of molecules of the gas

A
becomes half

B
becomes two times

C
becomes 4 times

D
remains constant

Solution

From ideal gas equation, PV = nRT
This means $$n = \frac{PV}{RT}$$
Now, pressure and temperature both are doubled and volume is halved
So, clearly the number of moles will also become half
So, A is the correct answer.
Question 3
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The absolute temperature T of a gas is plotted against its pressure P for two different constant volumes V$$_{1}$$ and V$$_{2}$$ where V$$_{1}$$ > V$$_{2}$$. T is plotted along x-axis and P along y-axis.

A
Slope for curve corresponding to volume V$$_{1}$$ is greater than that corresponding to volume V$$_{2}$$.

B
Slope for curve corresponding to volume V$$_{2}$$ is greater than that corresponding to volume V$$_{1}$$.

C
Slope for both curves are equal.

D
Slope for both curves are unequal such that they intersect at T = 0.

Solution

From ideal gas equation, PV = nRT
This means that $$P = \frac{nR}{V}T$$
So, the slope of P-T curve = $$\frac{nR}{V} \alpha \frac{1}{V}$$
this means that more slope => less volume
Since we are given that $$V_{2} < V_{1}$$, so we get that
Slope for curve corresponding to volume $$V_2$$ is greater than that corresponding to volume $$V_1$$.
So, B is the correct answer.
Question 4
1 / -0

A volume 'V' and temperature 'T' was obtained, as shown in the diagram, when a given mass of gas was heated. During the heating process, the pressure is

A
increased

B
decreased

C
remains constant

D
changed erratically

Solution

$$\frac {dV}{dT}= constant$$ from the plot.
But, $$\frac {dV}{dT}= T + \frac {dP}{dT}$$
$$ \therefore \frac {dP}{dT}=Constant -T$$
$$\therefore$$ P decreases as T increases.
Question 5
1 / -0

Universal gas constant per molecule is called

A
Rydberg constant

B
Kelvin constant

C
Boltzman constant

D
Stefan's constant
Question 6
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Equation of gas in terms of pressure $$p$$ absolute temperature $$T$$ and density $$d$$ is :

A
$$\dfrac{p_{1}}{d_{1}T_{1}}=\dfrac{p_{2}}{d_{2}T_{2}}$$

B
$$\dfrac{p_{1}T_{1}}{d_{1}}=\dfrac{p_{2}T_{2}}{d_{2}}$$

C
$$\dfrac{p_{1}d_{2}}{T_{2}}=\dfrac{p_{2}d_{1}}{T_{1}}$$

D
$$\dfrac{p_{1}d_{1}}{T_{1}}=\dfrac{p_{2}d_{2}}{T_{2}}$$

Solution

We know that the density of an ideal gas is given by the expression,

$$d = \dfrac{PM}{RT}$$ , where M is the molar mass of the gas

now, this means that $$\dfrac{P}{dT} = \dfrac{R}{M}$$

Now, both R and M are constant for a particular gas and do not depend on the physical parameters of the gas

So, $$\dfrac{P_{1}}{d_{1}T_{1}} = \dfrac{p_{2}}{d_{2}T_{2}}$$

So, A is the correct answer.
Question 7
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The density of air at N.T.P. is $$1.293 gm/lit$$. If the pressure is tripled keeping its temperature constant its density becomes

A
$$3.87 gm/ltr$$

B
$$1.293gm/ltr$$

C
$$2.586 gm/ltr$$

D
$$0.431 gm/ltr$$

Solution

First we derive an expression for density of a gas.
Density = $$\dfrac{mass(w)}{volume(V)}$$
but we know that PV = nRT
now n = $$\dfrac{w}{M}$$ where M is molecular weight
So, PV = $$\dfrac{w}{M}RT$$
this gives $$\dfrac{w}{V} = \dfrac{PM}{RT}$$ = density = d
Clearly at a constant temperature,
$$\dfrac{d_{1}}{P_{1}} = \dfrac{d_{2}}{P_{2}}$$
so since pressure is tripled, density also triples
new density = old density x 3
$$= 1.293 \times 3$$
$$ = 3.87 gm/ltr$$
Question 8
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A gas is kept at 13$$^{0}$$C in a vessel. If the volume of the gas is kept constant and is heated, the pressure will be doubled to its initial pressure at a temperature

A
572 K

B
286 K

C
143 K

D
73 K

Solution

PV = nRT
since n and R and V are constant

$$P_{1}T_{2} = P_{2}T_{1}$$
here $$V_{1} = V_{2}$$
and $$T_{1} = 286$$
and $$P_{2} = 2P_{1}$$
and $$T_{2} = ?$$ (to be determined)
Putting all the values, we get $$T_{2}= \dfrac{P_2 T_1}{P_1}= \dfrac{2 P_1 \times 286}{P_1}= 572K$$
Question 9
1 / -0

A bubble rises from the bottom of a lake, 90m deep. On reaching the surface, its volume becomes(take atmospheric pressure equals to 10 m of water )

A
4 times

B
8 times

C
10 times

D
3 times

Solution

At the bottom, pressure = 90m + 10m (atmospheric pressure) = 100m of water
at the top, pressure = 10m (only atmospheric pressure) = 10m of water
Since pressure has become one-tenth, volume becomes 10 times (Boyle's law)
Question 10
1 / -0

A cylinder contains a gas at temperature of 27$$^{0}$$C and a pressure 1MPa. If the temperature of the gas is lowered to -23$$^{0}$$ C , the change in pressure is

A
1MPa

B
5/6MPa

C
1/6MPa

D
5MPa

Solution

PV = nRT
Since n and R are constant, we get
$$\frac{PV}{T}$$ = constant
so, $$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$$
Here,
$$P_{1} = 1MPa$$
$$T_{1} = 300K$$
$$P_{2} = ?$$
$$V_{2} = V_{1}$$
$$T_{2} = 250K$$
Putting all the values, we get : $$P_{2} = \frac{5}{6}MPa$$
So change in pressure $$\frac{1}{6}$$MPa

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Thermal Properties of Matter Test - 24

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Thermal Properties of Matter Test - 24

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Question 1

is directly proportional to its pressure and absolute temperature

is directly proportional to its pressure and inversely proportional to its absolute temperature

is inversely proportional to its pressure and directly proportional to its absolute temperature

is inversely proportional to both its pressure and absolute temperature of the gas

Solution

Question 2

becomes half

becomes two times

becomes 4 times

remains constant

Solution

Question 3

Slope for curve corresponding to volume V$$_{1}$$ is greater than that corresponding to volume V$$_{2}$$.

Slope for curve corresponding to volume V$$_{2}$$ is greater than that corresponding to volume V$$_{1}$$.

Slope for both curves are equal.

Slope for both curves are unequal such that they intersect at T = 0.

Solution

Question 4

increased

decreased

remains constant

changed erratically

Solution

Question 5

Rydberg constant

Kelvin constant

Boltzman constant

Stefan's constant

Question 6

$$\dfrac{p_{1}}{d_{1}T_{1}}=\dfrac{p_{2}}{d_{2}T_{2}}$$

$$\dfrac{p_{1}T_{1}}{d_{1}}=\dfrac{p_{2}T_{2}}{d_{2}}$$

$$\dfrac{p_{1}d_{2}}{T_{2}}=\dfrac{p_{2}d_{1}}{T_{1}}$$

$$\dfrac{p_{1}d_{1}}{T_{1}}=\dfrac{p_{2}d_{2}}{T_{2}}$$

Solution

Question 7

$$3.87 gm/ltr$$

$$1.293gm/ltr$$

$$2.586 gm/ltr$$

$$0.431 gm/ltr$$

Solution

Question 8

572 K

286 K

143 K

73 K

Solution

Question 9

4 times

8 times

10 times

3 times

Solution

Question 10

1MPa

5/6MPa

1/6MPa

5MPa

Solution

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