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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm. And a temperature of `27^(@)C`. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm. And its temperature drops to `17^(@)C`. Estimate the mass of oxygen taken out of the cylinder. `(R = 8.1 J "mole"^(-1) K^(-1)`, molecular mass of `O_(2) = 32 u)`. |
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Answer» Initially in the oxygen cylinder, `V_(1) = 30 litre = 30 xx 10^(-3) m^(3)`, `P_(1) = 15 atm = 15 xx 1.01 xx 10^(5) Pa, T_(1) = 27 +273 = 300 K` If the cylinder contains `n_(1)` mole of oxygen gas, then `P_(1) V_(1) = n_(1)RT_(1)` or `n_(1) =(P_(1)V_(1))/(RT_(1)) = ((15 xx 1.01 xx 10^(5))xx(30xx10^(-3)))/(8.3 xx 300) = 18.253` For oxygen moleculaer weight, `M= 32 g` Initial mass of oxygen in the cylinder cylinder, `m_(1) = n_(1) M= 18.253 xx 32 = 548.1g` Finally in the oxygen gas in the cylinder, let `n_(2)` moles of oxygen be left, Here, `V_(2) = 30 xx 10^(-3) m^(3) , P_(2) = 11 xx 1.01 xx 10^(5)Pa , T_(2) = 17 + 273 = 290 K` Now, `n_(2) = (P_(2)V_(2))/(RT_(2)) = ((11xx1.01 xx 10^(5))xx(30xx10^(-3)))/(8.3 xx 290) = 13.847` `:.` Final mass of oxygen gas in the cylinder , `m_(2) = 13.847 xx 32 = 453.1 g` `:.` Mass of the oxygen gas withdrawn = `m_(1)-m_(2) = 584.1 - 453.1 = 131.0 g`. |
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| 102. |
What is the nature of graph of PV versus P for a given mass of a gas at constant temperature? |
| Answer» The graph is a stright line parallel to pressure axis. | |
| 103. |
Given is the graph between `(PV)/T` and P for 1 gm of oxygen gas at two different temperatures `T_(1) and T_(2)` Fig. Given, density of oxygen `= 1.427 kg m^(-3)`. The value of `(PV)//(T)` at the point A and the relation between `T_(1) and T_(2)` are respectively : A. `0.256 JK^(-1) and T_(1) lt T(2)`B. `8.314 mol^(-1) K^(-1) and T_(1) gt T(2)`C. `0.256 JK^(-1) and T_(1) gt T(2)`D. `4.28 JK^(-1) and T_(1) lt T(2)` |
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Answer» Correct Answer - C `PV = nRT = m/M RT` Where m = mass of the gas and `m/M = n =` number of moles `(PV)/(T) = nR = a` constant for all value of P. That is why, ideally it is a straight line. `:. (PV)/T = (1 gm)/(32 gm) xx 8.31 J "mole"^(-1) K^(-1)` `= 0.256 JK^(-1)`. Also, `T_(1) gt T_(2)`. |
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| 104. |
Fig shows of `PV//T` versus P for `1.00 xx 10^(-3) kg` of oxygen gas at two different temperatures. (a) What does the dotted plot signify ? (b) Which is true : `T_(1) lt T_(2) or T_(2) lt T_(1) ?` ( c) What is the value of `PV//T` where the curves meet on the Y-axis ? (d) If we obtained similar plot for `1.00 xx 10^(-3) kg` of hydrogen, would we get the same value of `PV//T` at the point where the curves meet on the y-axis ? If not, what mass of hydrogen yield the same value of `PV//T` (for low pressure high temperature region of the plot) ? (Molecular mass of `H = 2.02 u`, of `O = 32.0 u, R = 8.31 J "mol"^(-1) K^(-1)` |
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Answer» (a) The dotted plot shows that `(PV)/(T) (= mu R)` is a constant quantity, independent of pressure P. This signifies the ideal gas behaviour. (b) The curve at temperature `T_(1)` is closer to the dotted plot than the curve at temperature `T_(2)`.As the behaviour of a real approaches the behaviour of a perfect gas when temperature is increased, therefore, `T_(1) gt T_(2)`. (c) Where the two curves meet, the value of `PV//T` on Y-axis is equal to `mu R`. As mass of oxygen gas = `1.00 xx 10^(-3)Kg = 1g :. (PV)/(T) = mu R = (1/32) xx 8.31 KJ^(-1) = 0.26 JK^(-1)` (d) If we obtained similar plots for `1.00 xx 10^(-3) kg` of hydrogen , we will not get the same value of `PV//T` at the point , where the curves meet on the Y- axis . This is because molecular mass of hydrogen is different form that of oxygen. For the same value of `(PV)/T` , mass of hydrogen required is obtained from `(PV)/(T) = mu R = m/(2.02) xx 8.31 = 0.26` `m=(2.02 xx 0.26)/(8.31) gram = 6.32 xxx 10^(-2) gram`. |
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| 105. |
A cylinder of fixed capacity 44.8 litre contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by `15.0^(@) C ? [R = 8.31 J mol^(-1) K(-1)]` |
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Answer» It is known that one mole of any ideal gas at NTP occupies a volume of 22.4 litre. As the cylinder has a fixed capacity of 44.8 litre, it must contain 2 moles of helium at NTP. For Helium (momoatomic gas), `C_(upsilon) = 3/2 R` `:.` Heat required `=n` umber o f moles `xx` molar sp. `h e a t xx r i s e i n t e m p e r a t u r e` `Q=2 xx 3/2 R xx 15 = 45 r = 45 xx 8.31 J` `= 373.95 J`. |
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| 106. |
Isothermal curves for a given mass of gas are shown at two different temperture `T_(1) and T_(2)` in Fig. State whether `T_(1) gt T_(2) or T_(2)gt T_(1)`. Justify your answer. |
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Answer» From ideal gas equation `PV = mu RT` `T= (PV)/(mu R)` As mass of gas is constant, `mu` is constant , R already a constant. `:. T prop PV` Since PV is greater for the curve at `T_(2)` than for the curve at `T_(1)`, therefore, `T_(2) gt T_(1)`. |
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| 107. |
On reducing the volume of the gas at constant temperature, the pressure of the gas increases. Explain on kinetic theory. |
| Answer» On reducing the volume, the space for the given number of molecules of the gas decreases, i.e. No.of molecules collide with the walls of the vessel per second and hence a large momentum is transferred to the walls per second. That is why the perssure of the gas increases. | |
| 108. |
Why cooling is caused by evaporation? |
| Answer» As explaind in problem 13, evaporation occurs on account of faster molecules escaping from the surface of the liquid. The liquid is, therefore, left with molecules having lower speeds . The decrease in the averaged speed of the molecules results in lowering the temperature. That is why cooling is caused by evaporation. | |
| 109. |
Explain the rise of temperature on heating on the basis of kinetic theory. |
| Answer» When heat is given to a gas, the rms velocity of the gas molecules increases. As `C prop sqrt(T)`, so temperature of the gas increases. | |
| 110. |
Explain the phenomenon of evaporation on the basis of kinetic theory. |
| Answer» According to Kinetic theory, molecules of liquid are in a state of continuous random motion. The molecules near the surface of liquid may have enough kinetic energy so as over come the intermolecular attraction of other molecules on teh surface and hence manage escape. Such molecules would move around freely in the space above the liquid. This is the phenomenon of evaporation which may occur at all temperature. | |
| 111. |
Who proposed a model for a gas for the kinetic theory of gases? |
| Answer» R. Clausius and J.C. Maxwell. | |
| 112. |
A flask contains argon and chlorine in the ratio 2:1 by mass. The temperature of the mixture is `27^(@)`. Obtain the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed of the molecules of two gases. Atomic mass of argon = 39.9 u, Molecular mass of chlorine = 70.9 u. |
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Answer» we know that average `K.E.//"molecure"` of any ideal gas `= 3.2 K_(B) T` . Where `K_(B)` is Boltzmann constant and T is temperature. It does not depend upon nature of the gas. (i) As argon and chlorine, both have the same temperature in flask, the ratio of average `K.E//"moluecule"` of the gases = `1:1`. (i) If m is mass of a molecule of the gas, then average `K.E//"molecule" =1/2 m upsilon_(rms)^(2)` `:. 1/2 m_(1)upsilon_(1)^(2) = 1/2 m_(2)upsion_(2)^(2) = 3/2 K_(B)T` or ` (upsion_(1)^(2))/(upsion_(2)^(2)) = (m_(2))/(m_1) = (M_2)/(M_1) = (70.9)/(39.9) = 1.77` where M represents the `"atomic" // "molecular"` mass of the gas (For argon, a molecule is just an atom of argon). `:. (upsilon_1)/(upsilon_2) = sqrt(1.77) = 1.33` . |
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| 113. |
A cylinder has a mixture of hydrogen and oxygen gases in the ratio 1:3. the ratio of mean kinetic energies of the two gases is : |
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Answer» Correct Answer - 1 Mean kinetic energy of gas molecules is a function of temperature only. As temp.of two gases is the same, ratio of their mean kinetic energies is 1. |
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| 114. |
A certain amount of gas is sealed in a glass flask at 1 atmosphere pressure and `20^(@)C`. The flask can withstand upto a pressure of 2 atmosphere. Find the temperature to which gas can be heated so that the flask does not break.A. `513^(@)C`B. `413^(@)C`C. `313^(@)C`D. `213^(@)C` |
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Answer» Correct Answer - C Here, `P_(1) = 1 atm, P_(2) = 2 atm, T_(1) = 20^(@)C, T_(2)=?` As volume remains constant `:. (P_1)/(T_1) = (P_2)/(T_2)` `:. T_(2) = T_(1)(P_2)/(P_1) = (20+273) xx 2/1 = 293 xx 2 = 586K` `= 586 - 273 = 313^(@)C`. |
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| 115. |
A cylinder contains He at 2 atmosphere pressure and hydrogen at 1.5 atmosphere pressure. What is the pressure of the mixture of gases in the cylinder ? |
| Answer» `P= p_(1)+p_(2) =(2+1.5) atm = 3.5 atm`. | |
| 116. |
Two vessel separately contains two ideal gases A and B at the same temperature, the pressure of A being twice that of B. under such conditions, the density of A is found to be 1.5 times the density of B. the ratio of molecular weight of A and B isA. `1//2`B. `2//3`C. `3//4`D. 2 |
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Answer» Correct Answer - C Here, `rho_(A) = 1.5 rho_(B)` or `rho_(B)` . Also `P_(A) = 2 rho_(B)` As, `P = (RT rho)/(M) :. P_(A) = (Rtrho_(A))/(M_A), P_(B) = (RT rho_(B))/(M_B)` `(P_A)/(P_B) = (rho_(A) xx M_(B))/(rho_(B) xx M_(A)) or (M_A)/(M_B) = (rho_A)/(rho_B) xx (P_B)/(P_A)` `(M_A)/(M_B) = 3/2 xx 1/2 = 3/4`. |
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| 117. |
At what temperature will the average velocity of oxygen molecules be sufficient to escape from the earth. Given mass of oxygen molecule `= 5.34 xx 10^(-26) kg`. Boltzmann constant, `k = 1.38 xx 10^(-23) J "molecule"^(-1) K^(-1)`. Escape velocity of earth `= 11.0 km s^(-1)`. |
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Answer» Correct Answer - `1.56 xx 10^(5)K` Here, `T=?` For oxygen molecule to escape from the earth, average KE of molecule = Escape energy of molecule `3/2 kT = 1/2 m upsilon_(e)^(2)` `T = (m upsilon_(e)^(2))/(3k) = (5.34 xx 10^(-26)(11 xx 10^(3))^(2))/(3 xx 1.38 xx 10^(-23))` `=1.56 xx 10^(5)K` |
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| 118. |
Oxygen and hydrogen gas are at same temperature and pressure. And the oxygen molecule has 16 times the mass of hydrogen molecule. Then the ratio of their rms speed isA. 2B. `1//4`C. 4D. 16 |
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Answer» Correct Answer - B Here, `m_(O) = 16 m_(H)` Also, `C_(rms) prop 1/(sqrt(m)) :. (C_o)/(C_H) = sqrt((m_H)/(m_O))` or `(C_O)/(C_H) = sqrt((m_H)/(16m_H)) = 1/4`. |
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| 119. |
Four cylinders contain equal number of moles of argon, hydrogen, nitrogen and carbon dioxide at the same temperature. The energy is minimum inA. argonB. carbon dioxideC. nitrogenD. hydrogen |
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Answer» Correct Answer - A `KE//"mole"` at a particular temperature is minimum for a monoatomic gas `:.` it is minimum for argon. |
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| 120. |
At standard temperature and pressure, the mean free path of He gas is 300 mm. (a) Determine the effective diameter of He atom (b) the number of atoms per cubic metre. |
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Answer» Correct Answer - `1.67 xx 10^(10)m, 2.69 xx 10^(25)m^(-3)` Here `T = 0^(@)C = 273 K`, `P = 1 atm = 1.013 xx 10^(5) Nm^(-2)` `lambda = 300 nm = 3xx10^(-7) m, d = ? , n=?` (a) As `lambda = (kT)/(sqrt(2) pi d^(2)P)` `:. D = [(kT)/(sqrt(2) pi lambda P)]^(1//2` `=[(1.38 xx 10^(-23 xx 273))/(1.414 xx 3.14 xx 3 xx 10^(-7) xx 1.013 xx 10^(5))]^(1//2)` `=1.67 xx 10^(10)m` (b) From `lambda = (1)/(sqrt(2)pi d^(2)n)` `n = (1)/(sqrt(2)pi d^(2)n)` `(1)/(1.414 xx 3.14(1.67 xx 10^(-10))^(2) xx 3 xx 10^(-7))` `n=2.69 xx 10^(25)m^(-3)`. |
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| 121. |
Let `barv,v_(rms) and v_p` respectively denote the mean speed. Root mean square speed, and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m. ThenA. `(upsilon)_(P) lt vec(upsilon) lt (upsilon)_(rms)`B. no molecule can have a speed greater than `sqrt(2) (upsilon)_(rms)`C. no molecule can have speed less than `(upsilon)_(P)//(sqrt(2))`D. the average kinetic energy of a molecule is `(3)/(4)m (upsilon)^(2)_(P)` |
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Answer» Correct Answer - A::D Mean speed, `bar(upsilon) = sqrt((8kT)/(pi m)) = 0.92 upsilon_(rms)` rms speed, `upsilon_(rms) = sqrt((3kT)/(m))` most probable speed, `upsilon_(p)= sqrt((2kT)/(m)) = 0.816upsilon_(rms)` ltbr. As is clear from the formulae, `upsilon_(p) lt bar(upsilon) lt upsilon_(rms)` Average kinetic energy of a molecule `=3/2 kT = 3/4 m. ((2kt)/(m)) = 3/4 m upsilon_(p)^(2)`. |
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| 122. |
Uranium has two isotopes of masses 235 and 238 units. If both are present in uranium hexa fluoride gas, which would have the larger average speed ? If atomic mass of fluorine is 19 units, estimate the percentage difference in speed at any temperature. |
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Answer» At a fixed temperature , average `K.E.//"molecul e"= 1/2 m upsilon^(2)= 3/2 K_(B)T` = constat `:. 1/2 m_(1) upsilon_(1)^(2) = 1/2 m_(2) upsilon_(2)^(2) or (upsilon_1)/(upsilon_2) = sqrt((m_2)/(m_1))` Clearly, smaller the mass of the molecule, larger will be its average speed. For Uranium (235) hexafluoride, `m_(1)= 235 + 6 xx 19 = 349` units and for Uranium (238) hexafluoride, `m= 238 + 6 xx 19 = 352` units `U-235` hexafluoride will have larger average speed. `:. (upsilon_1)/(upsilon_2) = sqrt((352)/(349)) = 1/0044` `%` age difference `=(1.0044 - 1)/(1) xx 100` `= 0.44%`. |
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| 123. |
(a) When a molecule (or an elastic ball) hits a (massive) wall, it rebounds with the same speed. When a ball hits a massive bat held firmly, the same thing happens However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower? (b) When gas in a cylinder is compressed by pushing in a piston. Its temperature rises. Guess at an explanation of this in terms of kinetic theory using (a) above (c) What happens when a compressed gas pushes a piston out and expands. What would you observe? (d) Sachin Tendulkar uses a heavy cricket bat while playing. Does it help him in any way? |
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Answer» Suppose the speed of the ball relative to the wicket behind the bat is u. If the bat is moving towards the ball with a speed V relative to the wicket , then the relative speed of the ball w.r.t. `Bat = (V+u)`, towards the bat. On hitting the massive bat, the ball rebounds with the same speed `(V+u)`, away from the bat. Therefore, speed of rebounding ball relative to wicket `=V+(V+u) = 2V+u`, away from the wicket. Thus, the ball speeds up (moves faster) after the collision with the bat. for a molecule, it implies increase in temperature. Thus, cylinder acts as wicket , piston acts as bat and molecule act as ball. (b) When gas in a cylinder is compressed by pushing in a piston, bat hits the ball, it moves faster and hence the temperature of the gas increases. (c) When a compressed gas pushes a piston out and expand , velocity of ball decrease. hence temperature of the gas decreases. (d) Sachin Tendulkar uses a heavy cricket bat. Therefore, on hitting with such a bat , velocity of the ball increases further. He is able to score better. |
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| 124. |
Calculate the temperature at which rms velocity of a gas is one third its value at `0^(@)C`, pressure remaining constant. |
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Answer» As `C/(C_0) = sqrt((T)/(T_0)) =1/3` `:. T/(T_0) = 1/9 , T = (T_0)/(9) = (273)/(9) = 30.3K`. |
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| 125. |
A surface is hit elastically and normally by n balls per second, each of mass m moving with velocity u. if each ball is made to hit the same surface with velocity 2 u, the force on the surface would become K times. What is the value of K ? |
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Answer» Correct Answer - 2 Here, initial force, `F_(1)` = rate of change of linear momentum of balls `=nxxm(-u-u) = -2nm u` Now force, `F_(2) = n xxm(-2 u - 2 u)` `=-4 nm u = 2 F_(1)` `:. K=2`. |
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| 126. |
If masses of all molecules of a gas are halved and the speed doubled. Then the ratio of initial and final pressure is : |
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Answer» Here, `M_(1)=M, C_(1)=C, V_(1)=V, P_(1) =?, M_(2)=M//2, C_(2)=2 C, V_(2)=V, P_(2)=?` As `P_(1)=1/3 (M_1)/(V_1) C_(1)^(2)` and `P_(2)= 1/3 (M_2)/(V_2) C_(2)^(2) :. (P_1)/(P_2) = (M_1)/(M_2) xx (V_2)/(V_1) xx (C_(1)^(2))/(C_(2)^(2)) = (M)/((M//2)) xx V/V xx ((C)/(2C))^(2)`. |
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| 127. |
ABCDEFGH is hollow cube made of an insulator, Fig. Face ABCD has positive charge on it. Inside the cube, we have ionized hydrogen. The usual kinetic theory expression for pressure A. will be validB. will not be valid since the ions would experience forces other than due to collisions with the walls.C. will not be valid since collisions with walls would not be elastic.D. will not be valid because isotropy is lost. |
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Answer» Correct Answer - B::D As face ABCD has positive charge on it and the gas consists of ironzed hydrogen, therefore, isotorpy is lost. The usual expression for pressure on the basis of kinetic theory will not be valid would also experience forces other than the force due to collison with the walls of the container. Choices (b) and (d) are correct. |
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| 128. |
There are N molecules of a gas in a containter. If this number is increased to 2N, what will be (i) pressure (ii) total energy (iii) rms speed of the gas ? |
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Answer» (i) As `P=1/3 M/V c^(2) = 1/3 (mN)/(V) C^(2)`, therefore, when N is doubled, P becomes twice. (ii) As `E=3/2 P`, therefore, when P is doubled, total energy E is also doubled. (iii) rms speed remains the same as it depends only on temperature. |
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