Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 9

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Question 1
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Three perfect gases at absolute temperatures $$\mathrm{T}_{1},\ \mathrm{T}_{2}$$ and $$\mathrm{T}_{3}$$ are mixed. The masses of molecules are $$\mathrm{m}_{1},\ \mathrm{m}_{2}$$ and $$\mathrm{m}_{3}$$ and the number of molecules are $$\mathrm{n}_{1},\ \mathrm{n}_{2}$$ and $$\mathrm{n}_{3}$$ respectively. Assuming no loss of energy, the final temperature of the mixture is:

A
$$\displaystyle \frac{(\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3})}{3}$$

B
$$\displaystyle \frac{\mathrm{n}_{1}\mathrm{T}_{\mathrm{l}}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}$$

C
$$\displaystyle \frac{\mathrm{n}_{1}\mathrm{T}_{1}^{2}+\mathrm{n}_{2}\mathrm{T}_{2}^{2}+\mathrm{n}_{3}\mathrm{T}_{3}^{3}}{\mathrm{n}_{1}\mathrm{T}_{1}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}$$

D
$$\displaystyle \frac{\mathrm{n}_{1}^{2}\mathrm{T}_{1}^{2}+\mathrm{n}_{2}^{2}\mathrm{T}_{2}^{2}+\mathrm{n}_{3}\mathrm{T}_{3}^{3}}{\mathrm{n}_{1}\mathrm{T}_{1}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}$$

Solution

Number of moles of first gas $$=\displaystyle \frac{\mathrm{n}_{1}}{\mathrm{N}_{\mathrm{A}}}$$
Number of moles of second gas $$=\displaystyle \frac{\mathrm{n}_{2}}{\mathrm{N}_{\mathrm{A}}}$$
Number of moles of third gas $$=\displaystyle \frac{\mathrm{n}_{3}}{\mathrm{N}_{\mathrm{A}}}$$
If no loss of energy then
$$\mathrm{P}_{1}\mathrm{V}_{1}+\mathrm{P}_{2}\mathrm{V}_{2}+\mathrm{P}_{3}\mathrm{V}_{3}=$$ $$PV$$

$$\displaystyle
\frac{\mathrm{n}_{1}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{1}+\frac{\mathrm{n}_{2}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{2}+\frac{\mathrm{n}_{3}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{3}
$$ $$\displaystyle
=\frac{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{mix}
$$

$$\implies \displaystyle \mathrm{T}_{mix}=\frac{\mathrm{n}_{1}\mathrm{T}_{1}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}$$.
Question 2
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An ideal gas goes through a reversible cycle $$a\rightarrow b\rightarrow c\rightarrow d$$ has the V- T diagram shown below. Process $$d\rightarrow a $$ and $$b\rightarrow c$$ are adiabatic.
The corresponding P-V diagram for the process is ( all figure are schematic and not drawn to scale).

A

B

C

D

Solution

In process $$a \rightarrow b$$ and $$c \rightarrow d$$ the volume is directly proportional to temperature of gas, hence the process is isobaric, however, the slope of curve is different in each case. using ideal gas equation
$$PV=nRT \Rightarrow V=\dfrac{nR}{P} T$$
slope $$m=\dfrac{nR}{P}$$ for a high pressure the slope of the V-T curve will be less. In given diagram the slope of $$c \rightarrow d$$ process is lower than $$a \rightarrow b$$ process, hence process $$a \rightarrow b$$ is at higher pressure, from which option B and D can be ruled out.
Now initial rate of change of volume w.r.t. temperature is very less from $$b \rightarrow c$$ and $$d \rightarrow a$$ as shown in V-T diagram.
hence slope of P-V curve should tend to $$\lim_{\Delta V\rightarrow 0}\dfrac{\partial P }{\partial v}=\infty$$ as initial change in volume is nearly zero, which can be verified in option A
Hence correct answer is option A.
Question 3
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A leak proof cylinder of length $$1m$$, made of a metal which has low coefficient of expansion is floating vertically in water at $$0^oC$$ such that its height above the water surface is $$20cm$$. When the temperature of water is increased to $$4^oC$$, the height of the cylinder above the water surface becomes $$21cm$$. The density of water at $$T = 4^oC$$, relative to the density at $$T = 0^oC$$ is close to:

A
$$1.26$$

B
$$1.03$$

C
$$1.04$$

D
$$1.01$$

Solution

Let $$A$$ be the area of cross section of the cylinder.

$$p_0$$ be the density of water at $$0^oC$$

$$p_4$$ be the density of water at $$4^oC$$.

Total length of the cylinder is $$100cm$$

Given that the length of the cylinder above the surface of water is $$20cm$$

At $$0^oC$$, the weight of the cylinder is balanced by the buoyant force.

$$p_0A \times 0.8 \times g = w$$ [Buoyant force $$= pA lg$$] ...(i)

At 4^oC$$, thef weight of the cylinder is balanced by the buoyant force.

$$p_4A\times 0.79\times g = w$$ ...(ii)

Dividing (i) by (ii)

$$\dfrac{p_0\times 0.8}{p_4 \times 0.79} = 1$$, $$p_4 = \dfrac{p_0\times 0.8}{0.79} = 0.012p_0$$

$$\dfrac{p_4}{p_0} = 1.012 \simeq 1.01$$
Question 4
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Which of the following shows the correct relationship between the pressure $$'P'$$ and density $$\rho$$ of an ideal gas at constant temperature ?

A

B

C

D

Solution

From ideal gas law:
$$PV=nRT$$
$$P=\cfrac{MnRT}{MV}, \quad M:$$ Molecular mass number
But $$\rho=Mn/V$$
$$P=\rho RT/M$$

At constant temperature, $$P \propto \rho$$
Hence, the correct graph is (d).
Question 5
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$$100$$ $$\mathrm{g}$$ of water is heated from $$30^{\mathrm{o}}\mathrm{C}$$ to $$50^{\mathrm{o}}\mathrm{C}$$. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is $$4184$$ $$ \mathrm{J}/\mathrm{K}\mathrm{g}/\mathrm{K}$$)

A
$$4.2$$ $$\mathrm{k}\mathrm{J}$$

B
$$8.4$$ $$\mathrm{k}\mathrm{J}$$

C
$$84$$ $$\mathrm{k}\mathrm{J}$$

D
$$2.1$$ $$\mathrm{k}\mathrm{J}$$

Solution

$$\Delta Q= \Delta U +\Delta W$$

Since $$\Delta W =0$$, $$\Delta Q= \Delta U $$

$$\Delta U=ms \Delta T= 0.1\times 4.184 \times 20=8.368 kJ$$
Question 6
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Two moles of an ideal gas with $$\dfrac{C_P}{C_V}=\dfrac{5}{3}$$ are mixed with $$3$$ moles of another ideal gas with $$\dfrac{C_P}{C_V}=\dfrac{4}{3}$$. The value of $$\dfrac{C_P}{C_V}$$ for the mixture is :

A
$$1.47$$

B
$$1.45$$

C
$$1.42$$

D
$$1.50$$

Solution

Let $$n_1,n_2$$ be the number of moles of gases $$1$$ and $$2$$.
We have $$\dfrac{C_{p_1}}{C_{v_1}}=\dfrac{5}{3}$$ and We know $$C_{p_1}=C_{v_1}+R$$

$$\therefore \dfrac{5C_{v_1}}{3}=C_{v_1}+R,\dfrac{2C_{v_1}}{3}R_1,C_{v_1}=\dfrac{3R}{2}$$

$$A$$ and $$C_{p_1}=C_{ v_ 1 }+R=\dfrac{5R}{2}$$

Also, $$\dfrac{C_{ p_ 2}}{C_{ v_ 2 }}=\dfrac{4}{3}$$ and

$$C_{p_2}=C_{v_2}+R$$

$$\therefore \dfrac{4C_{ v_ 2 }}{3}=C_{ v_ 2 }+R, \dfrac{C_{ v_ 2 }}{3}=R,C_{ v_ 2 }=3R$$

$$C_{ p_ 2}=C_{ v_ 2 }+ R=4R$$

Now $$C_{ v_ {mix} }=\dfrac{n_1C_{ v_ 1 }+n_2C_{ v_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{3R}{2}+3\times 3R}{2+3}$$

$$=\dfrac{3R+9R}{5}=\dfrac{12R}{5}$$

$$C_{ p_ {mix} }=\dfrac{n_1C_{ p_ 1 }+n_2C_{ p_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{5R}{2}+3\times 4R}{2+3}$$

$$=\dfrac{17R}{5}$$

$$l_{mix}=\dfrac{C_{ p_ {mix} }}{C_{ v_ {mix} }}=\dfrac{17}{12}=1.4166=1.42$$

Option (C) is correct.
Question 7
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Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2:3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4:3. The ratio of their densities is

A
1 : 4

B
1 : 2

C
6 : 9

D
8 : 9

Solution

$$
Sol$$ : (D)
$$
PV
=nRT
=\dfrac{m
}{M
}RT
$$
$$
$$
$$
\Rightarrow PM
=\rho RT
$$
$$
$$
$$
\dfrac{\rho
_{1}}{\rho
_{2}}=\dfrac{P
_{1}M
_{1}}{P
_{2}M
_{2}}=(\dfrac{P
_{1}}{P
_{2}})\times(\dfrac{M
_{1}}{M
_{2}})=\dfrac{4}{3}\times\dfrac{2}{3}=\dfrac{8}{9}
$$
Here $$\rho _{1}$$ and $$\rho _{2}$$ are the densities gases in the vessel containing the mixture.
Question 8
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A given sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule of the gas is m. which of the following gives the density of the gas?

A
mKT

B
P/(kT)

C
Pm/(kT)

D
P/(kTV)

Solution

Hint: Use ideal gas equation.

Step 1: Note the given values
Volume =V
Pressure =P
Temperature =T
Mass of each molecule =m

Step 2: Calculate density of gas using ideal gas equation .
From the ideal gas law,
$$PV=NkT$$
$$\implies \dfrac{N}{V}=\dfrac{P}{kT}$$

Density of gas= $$\dfrac {\text{Total mass} }{\text volume}$$
$$\rho= \dfrac{Nm}{V}=\dfrac{Pm}{kT}$$
$$\textbf{Hence option C correct}$$
Question 9
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A vessel of volume $$20\ L$$ contains a mixture of hydrogen and helium at temperature of $$27^{\circ}C$$ and pressure $$2\ atm$$. The mass of mixture is $$5\ g$$. Assuming the gases to be ideal, the ratio of mass of hydrogen to that of the helium in the given mixture will be

A
$$1 : 2$$

B
$$2 : 3$$

C
$$2 : 1$$

D
$$2 : 5$$

Solution

Let there are $$n_{1}$$ moles of hydrogen and $$n_{2}$$ moles of helium in the given mixture. As $$Pv = nRT$$
Then the pressure of the mixture
$$P = \dfrac {n_{1}RT}{V} + \dfrac {n_{2}RT}{V} = (n_{1} + n_{2}) \dfrac {RT}{V}$$
$$\Rightarrow 2\times 101.3\times 10^{3} = (n_{1} + n_{2}) \times \dfrac {(8.3\times 300)}{20\times 10^{-3}}$$
or, $$(n_{1} + n_{2}) = \dfrac {2\times 101.30 \times 10^{3} \times 20\times 10^{-3}}{(8.3)(300)}$$
or, $$n_{1} + n_{2} = 1.62 .... (1)$$
The mass of the mixture is (in grams)
$$n_{1}\times 2 + n_{2} \times 4 = 5$$
$$\Rightarrow (n_{1} + 2n_{2}) = 2.5 ..... (2)$$
Solving the eqns. $$(1)$$ and $$(2)$$, we get
$$n_{1} = 0.74$$ and $$n_{2} = 0.88$$
Hence, $$\dfrac {m_{H}}{m_{He}} = \dfrac {0.74\times 2}{0.88\times 4} = \dfrac {1.48}{3.52} = \dfrac {2}{5}$$.
Question 10
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A hot and a cold body are kept in vacuum separated from each other. Which of the following causes decrease in temperature of the hot body?

A
Radiation

B
Convection

C
Conduction

D
Temperature remains unchanged

Solution

Heat flow through vacuum is possible in radiation mode due to which temperature of hot body falls.