Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 62

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Weekly Quiz Competition

Question 1
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Choose the correct alternative:
Water expands on reducing its temperature below _______ $$^{\circ} C $$ .

A
$$0$$

B
$$4$$

C
$$8$$

D
$$12$$

Solution
Question 2
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The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $$K$$ and $$2K$$ and thickness $$x$$ and $$4x$$ respectively are $$T_{2}$$ and $$T_{1}$$ ($$T_{2} > T_{1}$$). The rate of heat transfer through the slab in a steady state is $$\displaystyle \left [\frac{A(T_{2}-T_{1})K}{x} \right ]f$$, where $$f$$ equals to :

A
$$1$$

B
$$\dfrac{1}{2}$$

C
$$\dfrac{2}{3}$$

D
$$\dfrac{1}{3}$$

Solution

Let T be the temperature of the interface.
$$\dfrac{dQ}{dt}=KA\dfrac{dT}{dx}$$
Since the rate of flow of heat across the interface is same,
$$KA\dfrac {(T_2-T)}{x}=2KA \dfrac{T-T_1}{4x}$$

$$ \Rightarrow T=\dfrac{2T_2+T_1}{3}$$
Substituting the value of T in $$\dfrac{dQ}{dt}$$ for slab of thickness
$$4x=2K\times A(\dfrac{T-T_1}{dx})$$

$$=2KA\times \dfrac{( \dfrac{2T_2+T_1}{3}-T_1)}{4x}$$

$$=\dfrac{KA}{3x} \times (T_2-T_1)$$
Comparing this with the given expression
$$\dfrac{KA}{x} \times (T_2-T_1)f$$
We get $$f=\dfrac{1}{3}$$
Question 3
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One end of a thermally insulated rod is kept at a temperature $$T_{1}$$ and the other at $$T_{2}$$. The rod is composed of two sections of lengths $$l_{1}$$ and $$ l_{2}$$ and thermal conductivities $$K_{1} $$ and $$ K_{2}$$ respectively. The temperature at the interface of two sections is :

A
$$\displaystyle \frac{K_{2}l_{1}T_{1}+K_{1}l_{2}T_{2}}{K_{2}l_{1}+K_{1}l_{2}}$$

B
$$\displaystyle \frac{K_{1}l_{2}T_{1}+K_{2}l_{1}T_{2}}{K_{1}l_{2}+K_{2}l_{1}}$$

C
$$\displaystyle \frac{K_{1}l_{1}T_{1}+K_{2}l_{2}T_{2}}{K_{1}l_{1}+K_{2}l_{2}}$$

D
$$\displaystyle \frac{K_{2}l_{2}T_{1}+K_{1}l_{1}T_{2}}{K_{1}l_{1}+K_{2}l_{2}}$$

Solution

Let $$T$$ be the temp at the interface.
The rate of flow of heat remains same across the interface.
The equation for heat flow in conduction is:
$$ \dfrac {dQ}{dT} = -KA\dfrac {dT}{dx}$$ where, A is the cross section area across which heat flows.
$$ \Rightarrow K_1A \dfrac {(T_1-T)}{l_1} =K_2A \dfrac {(T-T_2)}{l_2} $$
$$ \Rightarrow \dfrac {(T_1-T)}{\dfrac {l_1}{K_1}} = \dfrac {(T-T_2)}{\dfrac {l_2}{K_2}}$$

$$ \Rightarrow \dfrac {T_1-T}{T-T_2}=\dfrac {l_1K_2}{l_2K_1}$$

$$ \Rightarrow T(K_1l_2+K_2l_1)= K_1l_2T_1+K_2l_1T_2$$

$$ \Rightarrow T= \dfrac {K_1l_2T_1+K_2l_1T_2}{K_1l_2+K_2l_1}$$
Question 4
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A closed container of volume 0.02 m$$^3$$ contains a mixture of neon and argon gases at a temperature of 27$$^{0}$$ C and at a pressure of $$1\times 10^{5}N/m^{2}$$. The total mass of the mixture is 28 g. If the gram molecular weights of neon and argon are 20 and 40 respectively, the masses of the individual gases in the container are respectively(assuming them to be ideal) [R = 8.314 J/mol K]

A
16 gm, 12 gm

B
4 gm, 24 gm

C
6 gm, 22 gm

D
12 gm, 16 gm

Solution

$$n=\frac { PV }{ RT } \\ n=\frac { 1\times { 10 }^{ 5 }\times 0.02 }{ 8.314\times 300 } \\ n=0.802\\ { n }_{ 1 }{ +n }_{ 2 }=0.802\\ { 20n }_{ 1 }{ +20n }_{ 2 }=16\quad \left( Multiplying\quad by\quad 20 \right) \\ { w }_{ 1 }{ +w }_{ 2 }=28\\ { n }_{ 1 }{ M }_{ neon }{ +n }_{ 2 }{ M }_{ argon }=28\\ 20{ n }_{ 1 }{ +40n }_{ 2 }=28\\ Solving\quad to\quad two\quad equations\quad we\quad get,\\ 20{ n }_{ 2 }=12\\ { n }_{ 2 }=0.6\\ \therefore { w }_{ 2 }=0.6\times 40=24gms\\ { w }_{ 1 }=28-24=4gms$$
Question 5
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A 90 cm long barometer tube contains some air above the mercury. The reading is 74.5 cm when the true pressure is 76 cm at the temperature $$15^{0} C$$. If the reading is observed to be 75.8 cm on a day when the temperature is $$5^{0}C$$, then the true pressure is:

A
77.38 cm of Hg

B
75.8 cm of Hg

C
74 cm of Hg

D
80 cm of Hg

Solution

Pressure = Actual pressure - Observed pressure
Volume = A(Barometer length - observed length)
$$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \\ \Rightarrow \dfrac { (76-74.5)(90-74.5) }{ 288 } =\dfrac { (P-75.8)(90-75.8) }{ 278 } \\ \Rightarrow P=77.38 \ \ cm \ \ of \ Hg$$
Question 6
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A smooth vertical tube with two different cross-sections is open at both ends. They are fitted with pistons of different areas of cross- section and each piston moves within a particular section. One mole of a gas enclosed between the pistons which are tied with non-stretchable threads. The difference in cross-sectional area of pistons $$10\mathrm{c}\mathrm{m}^{2}$$ The mass of gas confined between pistons is 5kg. The outside pressure is 1 atmosphere$$=10^{5}N/m^{2}$$. By how many degrees must the gas between pistons be heated to shift the piston by 5 cm?
Given $$R=8.3$$.

A
$$0.9^{\mathrm{o}}\mathrm{C}$$

B
$$1.2^{\mathrm{o}}\mathrm{C}$$

C
$$1.5^{\mathrm{o}}\mathrm{C}$$

D
$$0.5^{\mathrm{o}}\mathrm{C}$$

Solution

Let us consider $$s_1$$ and $$ s_2$$ as the cross-section area of the lower and upper piston respectively.
Let $$P_o$$ be the atmospheric pressure and P be the gas pressure.At equilibrium, the downward force and upward forces balances,

i.e., $$P_o + Ps_1 + mg = Ps_2 + P_{o}s_{1} $$ ($$Force = Pressure \times Area$$}

$$ P = P_{o} + \dfrac{mg}{s_1 - s_2}$$ -- (1)

Let $$l_1$$ and $$l_2$$ be the length of string in lower and upper section.
Initial volume of the gas = $$ l_1 s_1 + l_2 s_2$$

When the pistons shifts by a distance $$x$$,

Final volume = $$(l_1 - x)s_1 + (l_2 + x)s_2$$

We have $$ PV = RT$$. Applying this before heating and after heating the gas,

$$P(l_1 s_1 + l_2 s_2) = RT$$ -- (2)

$$P[(l_1 -x)s_1 + (l_2 + x)s_2] = R (T + \Delta T)$$

$$P[(l_1 s_1 + l_2 s_2)+ (s_2 - s_1)x]= R (T + \Delta T)$$

$$[P(l_1 s_1 + l_2 s_2)+ P(s_2 - s_1)x]= R (T + \Delta T)$$

$$P(l_1 s_1 + l_2 s_2)+ P \Delta S x= R (T + \Delta T)$$

Substituting for $$P(l_1 + l_2)$$ from equation 2,

$$RT+ P \Delta S x= R (T + \Delta T)$$

Substituting for $$P$$ from equation 1,

$$RT+ P_{o} + \dfrac{mg}{s_1 - s_2} \Delta S x= R (T + \Delta T)$$

$$\implies \Delta T = \dfrac{1}{R} [P_o + mgx)$$

$$ = \dfrac {1}{8.3} [10^5 \times 5 \times 10^{-2} \times 10 \times 0^{-4} + 5 \times 9.81 \times 5 \times 10^{-2} ]$$

$$ = 0.9$$
Question 7
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A horizontal uniform glass tube of 100cm length is sealed at both ends contains 10 cm mercury column in the middle the temperature and pressure of air on either side of mercury column are respectively 31$$^{0}$$C and 76cm of mercury if the air column at one end is kept at 0$$^{0}$$C and the other end at 273$$^{0}$$C the pressure of air which is at 0$$^{0}$$C is (in cm of Hg )

A
$$76$$

B
$$88.2$$

C
$$102.4$$

D
$$122$$

Solution

Let $$x$$ be the displacement of the mercury column on the side whose temperature is changed to $$0^{\circ}C$$. New pressure on both sides must be same so that the mercury column remains in equilibrium.
$$\dfrac{PV}{T}$$ is constant for air column.
$$\implies \dfrac{P_{0}(45)}{273+31}=\dfrac{P_1 (45-x)}{273+0}$$

and $$\dfrac{P_0 (45)}{273+31}=\dfrac{P_1(45+x)}{273+273}$$

Therefore, $$x=15\ cm$$ and thus $$P=102.4\ cm\ of\ Hg$$
Question 8
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A closed hollow insulated cylinder is filled with gas at 0$$^{0}$$C and also contains an insulated piston of negligible weight and negligible thickness at the the middle point. The gas at one side of the piston is heated to 100$$^{0}$$C . If the piston moves 5cm, the length of the hollow cylinder is

A
13.65 cm

B
27.3 cm

C
64.6 cm

D
54.6 cm

Solution

$$\dfrac{PV}{T}$$ is constant for gas in both sections of the cylinder.

For section with constant temperature,
$$P_{0}V_{0}=P_1V_1$$
$$\implies P_ (A\dfrac{L}{2})=P_1(A(\dfrac{L}{2}-5))$$

Similarly for section whose temperature increased,
$$\dfrac{P_0V_0}{T_0}=\dfrac{P_1V_1}{T_1}$$
$$\implies \dfrac{P_0(A\dfrac{L}{2})}{273}=\dfrac{P_1(A(\dfrac{L}{2}+5))}{373}$$
$$\implies L=64.6cm$$
Question 9
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Two identical containers each of volume V$$_{0}$$ are joined by a small pipe. The containers contain identical gases at temperature T$$_{0}$$ and pressure P$$_{0}$$. One container is heated to temperature 2T$$_{0}$$ while maintaining the other at the same temperature. The common pressure of the gas is P and n is the number of moles of gas in container at temperature 2T$$_{0}$$.

A
$$P=2P_{0}$$

B
$$P=\dfrac{4}{3}P_{0}$$

C
$$n=\dfrac{2P_{0}V_{0}}{3RT_{0}}$$

D
$$n=\dfrac{3P_{0}V_{0}}{2RT_{0}}$$

Solution

Since the total number of moles in the system remains the same before and after heating,

$$\dfrac{P_0V_0}{T_0}+\dfrac{P_0V_0}{T_0}=\dfrac{PV_0}{T_0}+\dfrac{PV_0}{2T_0}$$

$$\implies P=\dfrac{4}{3}P_0$$
Question 10
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Heat energy always flows from a body at low temperature to a body at high temperrature

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution