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Thermal Properties Of Matter

Q: 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:

Pressure = Actual pressure - Observed pressureVolume = 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\$

Q: 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 :

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}\$

Q: Two plates \$A\$ and \$B\$ of equal surface area are placed one on top of the other to form a composite plate of the same surface area. The thickness of \$A\$ and \$B\$ are \$4.0\$cm and \$6.0\$cm respectively. The temperature of the exposed surface of plate \$A\$ is \$-10\$\$^{o}\$C and that of the exposed surface of plate B is \$10\$\$^{0}\$C. Neglect heat loss from the edges of the composite plate, the temperature of the contact surface is \$T_{1}\$ if the plates A and B are made of the same material and \$T_{2}\$ if their thermal conductivities are in the ratio \$2:3\$ then

We use the resistance approach for this problem.The heat resistance is given by,\$ R = \dfrac{L}{KA} \$If the two plates are made of the same plate,\$ \dfrac{{K}_{a} A ({T}_{1} + 10)}{4} = - \dfrac{{K}_{b} A ({T}_{1} - 10)}{6} \$Solving this for \$ {T}_{1} \$ gives, it equal to -2Similarly, if the two plates are made of materials with thermal conductivities in the ratio, 2:3,\$ \dfrac{{K}_{a} A ({T}_{1} + 10)}{4} = - \dfrac{{K}_{b} A ({T}_{1} - 10)}{6} \$With \$ \dfrac{{K}_{a}}{{K}_{b}} = \dfrac{2}{3} \$Solving this, gives \$ {T}_{2} = 0 \$

Q: 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]

\$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\$

Q: 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 :

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}\$

Q: 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\$.

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\$

Q: 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 )

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\$

Q: 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

\$\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\$

Q: 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}\$.

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\$

Physics
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JEE Mains JEE Advanced NEET BITSAT Easy Qs Med Qs Hard Qs

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 $1 5^{0} C$ . 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:

77.38 cm of Hg

75.8 cm of Hg

74 cm of Hg

80 cm of Hg

Hard

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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 $2 K$ and thickness $x$ and $4 x$ respectively are $T_{2}$ and $T_{1}$ ( $T_{2} > T_{1}$ ). The rate of heat transfer through the slab in a steady state is $[\frac{A ( T _{2} - T _{1} ) K}{x}] f$ , where $f$ equals to :

1

\frac{1}{2}

\frac{2}{3}

\frac{1}{3}

Hard

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Two plates $A$ and $B$ of equal surface area are placed one on top of the other to form a composite plate of the same surface area. The thickness of $A$ and $B$ are $4.0$ cm and $6.0$ cm respectively. The temperature of the exposed surface of plate $A$ is $- 10$ $^{o}$ C and that of the exposed surface of plate B is $10$ $^{0}$ C. Neglect heat loss from the edges of the composite plate, the temperature of the contact surface is $T_{1}$ if the plates A and B are made of the same material and $T_{2}$ if their thermal conductivities are in the ratio $2 : 3$ then

T_{1} = - 4^{0} C

T_{1} = - 2^{0} C

T_{2} = - 3^{0} C

T_{2} = 0^{0} C

Hard

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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 $1 \times 1 0^{5} N / m^{2}$ . 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]

16 gm, 12 gm

4 gm, 24 gm

6 gm, 22 gm

12 gm, 16 gm

Hard

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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 :

\frac{K _{2} l _{1} T _{1} + K _{1} l _{2} T _{2}}{K _{2} l _{1} + K _{1} l _{2}}

\frac{K _{1} l _{2} T _{1} + K _{2} l _{1} T _{2}}{K _{1} l _{2} + K _{2} l _{1}}

\frac{K _{1} l _{1} T _{1} + K _{2} l _{2} T _{2}}{K _{1} l _{1} + K _{2} l _{2}}

\frac{K _{2} l _{2} T _{1} + K _{1} l _{1} T _{2}}{K _{1} l _{1} + K _{2} l _{2}}

Hard

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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 $10 c m^{2}$ The mass of gas confined between pistons is 5kg. The outside pressure is 1 atmosphere $= 1 0^{5} N / m^{2}$ . By how many degrees must the gas between pistons be heated to shift the piston by 5 cm?
Given $R = 8.3$ .

0 . 9^{o} C

1 . 2^{o} C

1 . 5^{o} C

0 . 5^{o} C

Hard

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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 )

76

88.2

102.4

122

Hard

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Assertion

Though air is a bad conductor yet a body kept in air loses heat quickly.

Reason

The heat loss is mainly due to radiation.

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Both Assertion and Reason are incorrect

Hard

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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

13.65 cm

27.3 cm

64.6 cm

54.6 cm

Hard

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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 $_{0}$ .

P = 2 P_{0}

P = \frac{4}{3} P_{0}

n = \frac{2 P _{0} V _{0}}{3 R T _{0}}

n = \frac{3 P _{0} V _{0}}{2 R T _{0}}

Hard

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Thermal Properties Of Matter

Physics 2707 Views

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 150C. If the reading is observed to be 75.8 cm on a day when the temperature is 50C, then the true pressure is:

One end of a thermally insulated rod is kept at a temperature T1​ and the other at T2​. The rod is composed of two sections of lengths l1​ and l2​ and thermal conductivities K1​ and K2​ respectively. The temperature at the interface of two sections is :

Though air is a bad conductor yet a body kept in air loses heat quickly.

The heat loss is mainly due to radiation.

A closed hollow insulated cylinder is filled with gas at 00C 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 1000C . If the piston moves 5cm, the length of the hollow cylinder is

Physics
2707 Views

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 $1 5^{0} C$ . 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:

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 :