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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.
| 1. |
Caculate molecular K.E. of 1 g of helium at NTP. What will be its energy at `100^(@)C`? |
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Answer» Correct Answer - `8.51 xx 10^(2)J`, 11.63 xx 10^(2)J` At NTP , `P=1.013 xx 10^(5) N//m^(2), T_(0) = 273 K` `V = 22.4 litre = 22.4 xx 10^(-3) m^(3), M=4` As helium is monoatomic gas, therefore, K.E.//mole = `3/2 RT_(0) = 3/2 PV` `K.E.//gram =3/2 (PV)/(M)` `= 3/2 xx (1.013 xx 10^(5) xx 22.4 xx 10^(-3))/(4)` `E_(0) = 8.51 xx 10^(2)J` |
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| 2. |
A gaseous mixture consists of 16 g of helium and 16 g of oxygen. Find `gamma` for the mixture. |
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Answer» Correct Answer - `1.62` Number of moles , `n_(1) = 16/4 = 4 , n_(2) = 16/32 = 1/2` `He , C_(upsilon) = 3/2 R`, For oxygen , `C_(upsilon) = 5/2 R` `C_(upsilon)(mixture) = (4 xx 3/2 R + 1/2 xx 5/2 R)/(4 +1/2) = 29/18 R` `gamma(mix) = 1+(R)/(C_(upsilon)(mic)) = 1+ r/(29//18 R)` `=1+18/29 = 47/29 = 1.62` |
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| 3. |
The absolute temperature of a gas is made 4 time its initial value. What will be the change in rms velocity of its molecules ? |
| Answer» As `T prop C^(2)`, therefore, `sqrt(T) prop C`. When T is made 4 times , rms velocity C becomes `sqrt(4)` times, i.e., 2 times. | |
| 4. |
What is the ratio of rms velocity of the molecules of an ideal gas at 240 k and 60 K ? |
| Answer» `(C_1)/(C_2) = sqrt((T_1)/(T_2)) = sqrt((240)/(60)) = sqrt(4) = 2` . | |
| 5. |
1 mole of monoatomic and one mole of diatomic gas are mixed together. The value of `C_(upsilon)` isA. 2 RB. `(3//2) R`C. RD. `R//2` |
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Answer» Correct Answer - A `C_(V) = (n_(1)C_(V_1)+n_(2)C_(V_2))/(n_(1)+n_(2)) = (1 xx (3)/(2)R + 1 xx (5)/(2)R)/(1+1) = 2R`. |
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| 6. |
Assertion : For monoatomic gas, `R//C_(upsilon) = 0.67`. Reason : For a monoatomic gas `C_(upsilon) = (3)/(2) R`.A. If, both Assertion and reason are ture and the Reason is the correct explanation of the Assertion.B. If both, Assertion and reason are true but Reason is not a correct expationation of the Assertion.C. If Assertion is true but the Reason is false.D. If Both, Assertion and reason are false. |
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Answer» Correct Answer - A Here, both the assertion and reason are true and reason is correct explanation of assertion. As. `(R)/(C_(upsilon)) = 0.67 , gamma = 1+(R)/(C_upsilon) = 1.67` which is true for a monoatomic gas. |
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| 7. |
Assertion : The ratio `C_(P)// C_(upsilon)` for a diatomic gas is more than that for a monoatomic gas. Reason : The moleculess of a monoatomic gas have more degrees of freedom than those of a diatomic gas.A. If, both Assertion and reason are ture and the Reason is the correct explanation of the Assertion.B. If both, Assertion and reason are true but Reason is not a correct expationation of the Assertion.C. If Assertion is true but the Reason is false.D. If Both, Assertion and reason are false. |
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Answer» Correct Answer - D For a monoatomic gas, no of degree of freedom `n=3`, and for a diatomic gas, `n=5`. As, `(C_p)/(C_(upsilon))= gamma = 1+(2)/(n) :. ((C_p)/(C_(upsilon))_(mono) gt ((C_p)/(C_(upsilon))_(di)` Both the assertion and reason are false. |
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| 8. |
Assertion : The ratio `C_(P)//C_(upsilon)` is more for helium gas than for hydrogen gas. Reason : Atomic mass of helium is more than that of hydrogen.A. If, both Assertion and reason are ture and the Reason is the correct explanation of the Assertion.B. If both, Assertion and reason are true but Reason is not a correct expationation of the Assertion.C. If Assertion is true but the Reason is false.D. If Both, Assertion and reason are false. |
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Answer» Correct Answer - B Helium is monoatomic and hydrogen is diatomic. Helium has smaller number of degrees of freedom than hydrogen. So `C_(p)//C_(upsilon)` for `He` is more than that for hydrogen. Assertion is true. Reason is true but it is not a correct explanation of assertion. |
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| 9. |
What is the value of gas constant in cgs system for 1 gram or helium ? |
| Answer» `2.078 xx 10^(7) erg g^(-1)`^(@)C^(-1)`. | |
| 10. |
An inflated rubber balloon contains one mole of an ideal gas has a pressure p, volume V and temperature T. if the temperature rises to 1.1 T, and the volume is increased to 1.05 V, the final pressure will beA. `1.1p`B. pC. less than `p`D. between `p` and `1.1p` |
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Answer» Correct Answer - D Here, `p_(1) = p, upsilon_(1) =V , T_(1)=T, T_(2) = 1.1 T, upsilon_(2) = 1.05 V, p_(2)=?` from `(p_(2)upsilon_(2))/(T_2) = (p_(1)upsilon_(1))/(T_1)` `p_(2)=p_(1) (upsilon_(1))/(upsilon_2)*(T_2)/(T_1) = p xx V/(1.05 V) xx (1.1T)/(T) = (1.1p)/(1.05) = 1.05p`. |
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| 11. |
An air bubble of volume `1.0 cm^(3)` rises from the bottom of a take 40 m deep at a temperature of `12^(@) C`. To what volume does it grow when it reaches the surface, which is at a temperature of `35^(@) C`. ? Given `1 atm = 1.01 xx 10^(5) Pa`. |
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Answer» `V_(1) = 1.0 cm^(3) = 1.0 xx 10^(-6) m^(3) , T_(1) = 12^(@)C = 12+273 = 285 K`, `P_(1) = 1 at, + h_(1) rho g = 1.01 xx 10^(5) + 40 xx 10^(3) xx 9.8 = 493000 Pa`. when the air bubble reaches at the surface of lake , then `V_(2) = ? , T_(2) = 35^(@)C = 35 + 273 = 308 K, P_(2) = 1 atm , = 1.01 xx 10^(5) Pa` Now , `(P_(1)V_(1))/(T_1) = (P_(2)V_(2))/(T_2)` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))` `:. V_(2) = ((493000) xx (1.0xx10^(-6))xx308)/(285 xx 1.01 xx 10^(5)) = 5.275 xx 10^(-6)m^(3)`. |
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| 12. |
A gas enclosed in a vessel has pressure P, volume V and absolute temperature T, write the formula for number of molecule N of the gas. |
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Answer» From `PV = n RT = N/(N_A) RT` `N=(PVN_(A))/(RT) =(PV)/(kT)` where `k=R/(N_A)` = Boltzmann constant. |
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| 13. |
At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at `47^(@)C`? |
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Answer» Let `T` be the desired temperature. As rms speed of `H_(2)` mol = rms speed of oxygen mol `:. Sqrt((3RT_(H))/(M_H)) = sqrt((3RT_("oxy"))/(M_("oxy")))` or, `(T_H)/(2) = (273 + 47)/(32) , T_(H) = (320)/(16) = 20 K`. |
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| 14. |
If `C_(1), C_(2), C_(3)`......are random speed of gas molecules, then average speed `C_(av)= (C_(1)+C_(2)+C_(3)+...C_(n))/(n)` and root mean square speed of gas molecules, `C_(rms) = sqrt((C_(1)^(2)+C_(2)^(2)+C_(3)^(2)+...C_(n)^(2))/(n)) =C`. Further , `C_(2) prop T or C prop sqrt(T)` at `0k, C=0`, ie., molecular motion stops. With the help of the passage given above , choose the most appropriate alternative for each of the following question: Temperature of a certain mass of a gas is doubled. the rms speed of its molecules becomes n times. where n isA. `sqrt(2)`B. 2C. `(1)/(sqrt(2))`D. `1/2` |
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Answer» Correct Answer - A As, `C prop sqrt(T)` `:.` when T is doubled , C becomes `sqrt(2) times. |
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| 15. |
If three gas molecules have velocity 0.5, 1 and `2 km//s` respectively, find the ratio of their root mean square speed and average speed. |
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Answer» Here, `C_(1) = 0.5 km//s, C_(2) = 1 km//s` `C_(3) = 2 km//s` rms speed, `C= sqrt((C_(1)^(2)+C_(2)^(2)+C_(3)^(2))/(2))` `=sqrt(((0.5)^(2) + 1^(2)+2^(2))/(3)) = sqrt(1.75) = 1.323 km//s` `C_(av) = (C_(1)+C_(2)+C_(3))/(3) = (0.5 +1 +2)/(3) = 1.167 km//s` `(C)/(C_(av)) = (1.323)/(1.167) = 1.13`. |
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| 16. |
If three gas molecules have velocity `0.5`, `1` and `2 km//s` respectively, find the ratio of their root mean square speed and average speed. |
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Answer» Correct Answer - `1.134` r.m.s. Speed, `C_(rms) = sqrt((C_(1)^(2)+C_(2)^(2)+C_(3)^(2))/(3))` `sqrt(((0.5)^(2)+(1)^(2)+(2)^(2))/(3)) = 1.3228 km//s` Average speed, `C_(av) = (C_(1) +C_(2) +C_(3))/(3) = (0.5 +1 +2)/(3)` `= 1.167 km//s` `(C_(rms))/(C_(av))= 1.329/1.167 = 1.134` |
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| 17. |
At `27^(@)C`, a certain mass of gas exerts a pressure of 6 cm of Hg column. What will be the pressure (in cm of Hg) exerted when it is heated to `127^(@)C` at constant volume ? |
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Answer» Correct Answer - 8 Here, `T_(1)=27+273=300K` `P_(1)=6cm` of Hg col. `T_(2)=127+273 = 400K, P_(2)=?` as the gas is heated at constant volume, `(P_2)/(P_1) =(T_2)/(T_1) , P_(2) = P_(1)xx(T_2)/(T_1)` `=6xx400/300 = 8cm` of Hg col. |
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| 18. |
The pressure exerted by an ideal gas is `P = (1)/(3) (M)/(V)C^(2)`, where the symbols have their usual meaning. Using standard gas equation, PV = RT, we find that `C^(2) = (3 RT)/(M) or C^(2) oo T`. Average kinetic energy of translation of one mole of gas ` =(1)/(2) MC^(2) = (3 RT)/(2)` with the help of the passage given above, choose the most appropriate alternative for each of the following quetions : Average thermal energy of a helium atom at 600 K would beA. `6.21 xx 10^(-21) J`B. `1.24 xx 10^(-20) J`C. `1.24 xx 10^(-21) J`D. `1.24 xx 10^(21) J` |
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Answer» Correct Answer - B When `T = 600K` thermal energy//atom = `3/2 kT` `=3/2 xx (1.38 xx 10^(-23)) xx 600 = 1.24 xx 10^(-20)J`. |
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| 19. |
number of molecules in given mass m of substance = ………………………….. |
| Answer» `(6.023 xx 10^(23) xx m)/("molecularweight")` | |
| 20. |
Pressure exerted by a gas is due to …………….against…………………………… |
| Answer» continuous bombardment of gas molecules , the walls of the container. | |
| 21. |
A vessel of volume , `V = 5.0` litre contains `1.4 g` of nitrogen at a temperature `T = 1800 K`. Find the pressure of the gas if `30%` of its molecules are dissociated into atoms into atmos at this temperature. |
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Answer» The vessel contains a mixture of molecular nitrogen and atomic nitrogen . Let `m_(1), m_(2)` be the masses of teh molecular nitrogen and atomic nitrogen respectively. Then `m_(1) = 1.4 xx 70/100 = 0.98 g , m_(2) = 1.4 xx 30/100 = 0.42 g` Molecular wt. of nitrogen , `M_(1) = 28 , T= 1800 K, V= 5` litre `= 5 xx 10^(-3)m^(3)` Atomic wt. of nitrogen, `M_(2) = 14, R = 8.3 j "mole"^(-1)K^(-1)` Let `P_(1)` and `P_(2)` be the pressure due to molecular and atomic nitrogen respectively. therefore, `P_(1) = (mu_(1)RT)/(V) = (m_(1)RT)/(M_(1)V) = (0.98)/(28) xx (8.3 xx 1800)/(5 xx 10^(-3)) = 1.046 xx 10^(5) Nm^(-2)` `P_(2) (mu_(2)RT)/(V) = (m_(2))RT)/(M_(2)V) = 0.42/14 xx (8.3 xx 1800)/(5 xx 10^(-3)) = 0.894 xx 10^(5) Nm^(-2)` `:.` Net pressure of the gas = `P_(1) +P_(2) = 1.046 xx 10^(5) +0.894 xx 10^(5) = 1.94 xx 10^(5) Nm^(-2)` . |
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| 22. |
Calculate the total number of degree of freedom possessed by 10 c.c. of hydrogen gas at N.T.P |
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Answer» Correct Answer - `1.34375 xx 10^(21)` Number of hydrogen molecules in `22400` c.c. `= 6.02 xx 10^(23) at NTP` `:.` Number of hydrogen molecules in `10 c.c.` `=(6.02 xx 10^(23) xx 10)/(22400)` at NTP As hydrogen is diatomic gas, each molecule has 5 degree of freedom. `:.` Total no. of degree of freedom `= (5 xx 6.02 xx 10^(23) xx10)/(22400) = 1.34375 xx 10^(21)` |
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| 23. |
Calculate the number of degree of freedom in 15 c.c. Of nitrogen at N.T.P.? |
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Answer» `V=15 c.c. N=?` At N.T.P., number of nitrogen molecules in `22400 c.c. = 6.02 xx 10^(23)` `:.` number of nitrogen molecules in 15 c.c. `=(6.02 xx 10^(23) xx 15)/(22400)` As number of degrees of freedom of each diatomic mol. = 5 `:.` Total no. of degree of freedom `=(6.02 xx 10^(23) xx 15 xx 5)/(22400) = 2.015 xx 10^(21)`. |
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| 24. |
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume `V_1` and contains ideal gas at pressure `P_1` and temperature `T_1`.The other chamber has volume `V_2` and contains ideal gas at pressure `P_2` and temperature `T_2`. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will beA. `(T_(1) T_(2)(P_(1)V_(1) + P_(2)V_(2)))/(P_(1)V_(1)T_(2) + P_(2)V_(2)T_(1))`B. `(P_(1) V_(1)(T_(1) + P_(2)V_(2)T_(2)))/(P_(1)V_(1)+ P_(2)V_(2))`C. `(P_(1) V_(1)(T_(2) + P_(2)V_(2)T_(1)))/(P_(1)V_(1)+ P_(2)V_(2))`D. `(T_(1) T_(2)(P_(1)V_(1) + P_(2)V_(2)))/(P_(1)V_(1)T_(1) + P_(2)V_(2)T_(2))` |
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Answer» Correct Answer - A According to standard gas equation `(P_(1)V_(1))/(T_1) = mu_(1)R` and `(P_(2)V_(2))/(T_2) = mu^(2)R` ...(i) As no work is done in removing the partition, total energy remains conserved. Therefore, ltbr. `3/2 (P_(1)V_(1)+P_(2)V_(2)) = 3/2P(V_(1)+V_(2))` `:. P = (P_(1)V_(1)+P_(2)V_(2))/(V_(1)+V_(2))`..(ii) For mixuture of two gases `(mu_(1)+mu_(2))RT = P(V_(1)+V_(2))` Using (i) and (ii) `((P_(1)V_(1))/(RT_(1))+(P_(2)V_(2))/(T_2)) RT = ((P_(1)V_(1)+P_(2)V_(2))(V_(1)V_(2)))/((V_(1)+V_(2)))` `((P_(1)V_(1))/(T_(1))+(P_(2)V_(2))/(T_2))T = (P_(1)V_(1)+P_(2)V_(2))` On solving `T = ((P_(1)V_(1)+P_(2)V_(2))T_(1)T_(2))/((P_(1)V_(1)T_(1)+P_(2)V_(2)T_(2))`. |
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| 25. |
Assetion : Equal masses of helium and oxygen gases are given equal quantities of heat. The rise in temperature of helium is greater thant that in case of oxygen. Reason : The molecular mass of oxygen is more than molecular mass of helium.A. If, both Assertion and reason are ture and the Reason is the correct explanation of the Assertion.B. If both, Assertion and reason are true but Reason is not a correct expationation of the Assertion.C. If Assertion is true but the Reason is false.D. If Both, Assertion and reason are false. |
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Answer» Correct Answer - B Here, both assertion and reason are true but reason is not correct explanation of assertion. Heat given to helium is totally used up in increasing the translational energy of its molecules sime it is a monoatomic gas. On the other hand, in case of oxygen which is a diatomic gas, heat is used up in increasing transational , rotational and vibrational energies of its molecules. the rise in temperature of a gas is only, due to increase im transtaltional energy of its molecules. |
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| 26. |
Assetion : Specific heat of a gas at constant pressure is greater than its specific heat at constant volume. This is because at constant pressure, some heat is spent in expansion of the gas.A. If, both Assertion and reason are ture and the Reason is the correct explanation of the Assertion.B. If both, Assertion and reason are true but Reason is not a correct expationation of the Assertion.C. If Assertion is true but the Reason is false.D. If Both, Assertion and reason are false. |
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Answer» Correct Answer - A Reason is correct explanation of the assertion, which is true. |
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| 27. |
If typical size of a gas molecule is `2Å`, average distance between the molecules isA. `1 Å`B. `2 Å`C. `lt20 Å`D. `ge 20 Å` |
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Answer» Correct Answer - D `ge 20 Å` |
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| 28. |
In an isotopic gas, with no preferred direction of motion of molecules,A. `upsilon_(x) = upsilon_(y) = upsilon_(z)`B. `upsilon_(x)^(2) = upsilon_(y)^(2) = upsilon_(z)^(2)`C. `bar(upsilon_(x)^(2)) = bar(upsilon_(y)^(2)) = bar(upsilon_(z)^(2))`D. none of these |
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Answer» Correct Answer - C `bar(upsilon_(x)^(2)) = bar(upsilon_(y)^(2)) = bar(upsilon_(z)^(2))` |
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| 29. |
Which of the following processes are reversible ?A. DiffusionB. Change of stateC. ElectrolysisD. Heat conduction |
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Answer» Correct Answer - B::C Change of state and electrolysis, both are reversible, Diffusion and heat conduction are not. |
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| 30. |
Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature `T_0`, while Box contains one mole of helium at temperature `(7/3)T_0`. The boxes are then put into thermal contact with each other, and heat flows between them until the gasses reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gasses, `T_f` in terms of `T_0` isA. `T_(f) = (7)/(3)T_(0)`B. `T_(f) = (3)/(2)T_(0)`C. `T_(f) = (5)/(2)T_(0)`D. `T_(f) = (3)/(7)T_(0)` |
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Answer» Correct Answer - B Here, `n_(1) "mole" , n_(2) = 1 "mole"` For nitrogen, `C_(p_1)=7/2 R` for helium, `C_(p_2) = 5/2 R` `T_(1)=T_(0), T_(2) = 7/3 T_(0).T_(f)=?` when gases are put into thermal contact, heat is exchanged between them till final temperature `T_(f)` is reached. Heat gained by nitrogen = Heat lost by He `n_(1)C_(P_1)(T_(f)-T_(1)) = n_(2)C_(P_2) (T_(2)-T_(f))` `(n_(1)C_(P_1)+n_(2)C_(p_2)) T_(f) = n_(2)C_(p_2) T_(2) n_(1)C_(p_1)T_(1)` `(1 xx 7.2 R + 5/2 R) T_(f) = 1 xx 5/2 R xx 7/3 T_(0) + 1 xx 7/2 RT_(0)` `6RT_(f) = (35/6 + 7/2)RT_(0)` `T_(f) = (56)/(6 xx 6) T_(0)~~3/2 T_(0)`. |
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| 31. |
In the given (V-T) diagram, what is the relation between pressure `P_(1) and P_(2)` ? A. Cannot be predictedB. `P_(2) = P_(1)`C. `P_(2) gt P_(1)`D. `P_(2) lt P_(1)` |
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Answer» Correct Answer - D For an ideal gas, `PV = nRT or V = (nRT)/(P)` slope of V -T graph `= (dV)/(dT) = (nR)/(P)` `:.` Slope prop `1//P` As, `theta_(2) gt theta_(1)`, so `(1)/(P_2) gt (1)/(P_1) or P_(2) lt P_(1)`. |
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| 32. |
A nitrogen molecules at teh surface of earth happens to have the rms speed for that gas at `0^(@)`. If it were to go straight up without colliding with other molecules, how high would it rise? Mass of nitrogen molecules, `m = 4.65 xx 10^(26)lg, k=1.38 xx 10^(-23)J "molecule"^(-1) K^(-1)`. |
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Answer» Here, `k=1.38 xx 10^(-23) J "molecule" ^(-1)K^(-1), m=4.65 xx 10^(-26) kg , T = 0 +273 = 273 K` `upsilon_(rms) = sqrt((3kT)/(m)) = sqrt((3xx1.38 xx 10^(-23)xx273)/(4.65 xx 10^(-26))) = 493.1 m//s` The molecule goes to a height h till its entire K.E. is converted into P.E. `:. mgh = 1/2 m upsilon_(rms)^(2)` `h = (upsilon_(rms)^(2))/(2g) = ((493.1)^(2))/(2 xx 9.8) = 12.4 xx 10^(4) m = 12.4 km`. |
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| 33. |
Average straight distance covered between two successive collision of molecules is called…………………. |
| Answer» mean free path of molecules | |
| 34. |
Under what conditions do the real gases obey more strictly the gas equation, PV = RT ? Explain. |
| Answer» The basic properties of the molicules of an ideal gas are (i) zero size of the molecules and hence, zero volume of hte molecules and (ii) no mutual intermolecular force between them. At low pressure, the volume of the given gas becomes large. Therefore the volume of the molecules becomes negligible in comparison to the volume of the gas. At high temperature , the molecules have large K.E. and so the effect of the imtermolecular force on the motion of the pressure and high temperature, the real gases behave as ideal gases and gas equation is obeyed. | |
| 35. |
Can the temperature of a gas increased keeping its pressure and volume constant ? |
| Answer» No, temperature cannot be changed without changing v or P. | |
| 36. |
An ideal gas is taken from the state A (P, V) to the state B `(P//2, 2 V)` along a st. line path as shown in Fig. Select the correct statement from the following: A. work done by the gas in going from A to B exceed the work done in going from A to D under isothermal conditions,B. in the T - V diagram, part AB would become a parabolaC. in the T - V diagram, part AB would become a hyperbolaD. in going from A to D the Temp. Of gas first increases to a max value and then decreases |
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Answer» Correct Answer - A::B::D Isothermal curve from A to B will be parabolic with lesser area under the curve than the area under st. Line AB. Therefore, work done by the gas in going straight from A to B is more. `:.` Choice (a) is correct. If `P_(0), V_(0)` be the intercepts of curve on P and V axes, then its eqn. is obtained from = `y =mx + c` i.e., `P =(P_0)/(V_0) V+P_(0) or (RT)/(V) = (P_(0)V)/(V_0) +P_(0)` or, `T = (P_0)/(V_(0)R) V^(2)+(P_(0)V)/(R)` Which is the eqn. of a parabola. hence T-V curve is parabolic. therefore Choice (b) is correct. Also `(P//2)(2V) = PV` = const. i.e., process is isothermal. |
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| 37. |
Ten small planes are flying at a speed of `150 km//h` in total darkness in an air space that is `20 xx 20 xx 1.5 km^(3)` in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are, On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximately by a sphere of radius 10 m. |
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Answer» Time between near collision `t =(lambda)/(upsilon)` where `lambda` is mean free path = `(1)/(sqrt(2)pi n d^(2))` where d is diameter of safety region = `20 m = 20 xx 10^(-3) km` and n=number density ` N/V = (10)/(20 xx 20 xx1.5) = 0.0167 km^(-3)` we are given `upsilon=150 km//h` As `t = (lambda)/(upsilon) :. t = (1)/([sqrt(2) pi nd^(2)]upsilon) = 1/(1.414 xx 3.14 xx 0.0167 xx (20xx10^(-3))^(2) xx 150) = 225 hour`. |
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| 38. |
Why temperature less than OK is not possible ? |
| Answer» According to kinetic theory of gases, `T prop C^(2)` . At `T=0k, C^(2) = 0`, i.e. Molecular motion ceases, negative,therefore, temperature below 0K is not possible. | |
| 39. |
On driving the scooter for a long time, the air pressure in the tyres slightly increases. Why ? |
| Answer» On driving a scooter for a long time, the work done against friction is converted into heat. The gas in the tyre gets heated and hence the pressure of the gas increases, because `P prop T`. | |
| 40. |
The density of water is `1000 kg m^(-3)`. The density of water vapour at `100^(@) C` and 1 atmospheric pressure is `0.6 kg m^(-3)`. The volume of a molecule multiplied by the total number gives what is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure. |
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Answer» Here, density of water `= 1000 kg m^(-3)` density of water vapour `= 0.6 kg m^(-3)` For a given mass of water molecules, density is less if volume is large. `:. ("volume of water vap our")/("volume of water")` `=("density of water")/("density of water vapour")` `(1000)/(0.6) = (10^3)/(6 xx 10^(-1)) = (1)/(6 xx 10^(-4))` if density of bulk water and density of water molecule is same, then fraction of molecular volume to the total volume in liquid state `= 1`. As volume in vapour state has increased, therefore, fractional volume, i.e., molecular volume to the total volume is less by the same factor, i,e. `6 xx 10^(-4)`. |
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| 41. |
You are given the following group of particles `n_(1)` represents the number of molecules with speed `upsilone_(1)` |
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Answer» Correct Answer - `3.17 ms^(-1), 3.36 ms^(-1)`, `3.0 ms^(-1)` (i) The average speed is `upsilon_(av) = (n_(1)upsilon_(1)+n_(2)upsilon_(2)+n_(3)upsilon_(3)+n_(4)upsilon_(4)+n_(5)upsilon_(5))/((n_(1)+n_(2)+n_(3)+n_(4)+n_(5)))` `=(2 xx 1 +4 xx 2 + 8 xx 3 +6 xx 4 +3 xx 5)/((2+4+8+6+3)) = 73/223` `=3.17 ms^(-1)` (ii) The root mean square speed is , `upsilon_(rms) = ((n_(1)upsilon_(1)^(2)+n_(2)upsilon_(2)^(2)+n_(3)upsilon_(3)^(3)+n_(4)upsilon_(4)^(2)+n_(5)upsilon_(5)^(2))/(n_(1)+n_(2)+n_(3)+n_(4)+n_(5)))^(1//2)` `=((2 xx 1^(2) +4 xx 2^(2) + 8 xx 3^(2) +6 xx 4^(2) +3 xx 5^(2))/(2+4+8+6+3))^(1//2)` `=3.36 ms^(-1)` (iii) The most probable speed is that speed which is possessed by maximum number of particles, Accordingly , most probable speed = `3.0 ms^(-1)` |
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| 42. |
In Brownian movement, what should be the typical size of suspended particles and why ? |
| Answer» The typical size of suspended particles should be `~= 10^(-6)m`. If the size is too large, paricles would not move due to large inertia. If the size is too small, net momentum imparted to the suspended particle due to bambardment of neighbouring particles would be zero. | |
| 43. |
Atoms are not point masses. What is the simplest evidence in nature ? |
| Answer» Brownian motion is the simplest evidence. | |
| 44. |
Calculate the temperature at which rms velocity of a gas is half its value at `0^(@)C`, pressure remaining constant |
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Answer» Here, `T_(0)=0^(@)C = (0+273)K = 273 K` `T_(2) = ? upsilon_(2) = 1/2 upsilon_(1)` As , `(upsilon_2)/(upsilon_1) = sqrt((T_2)/(T_1)) = 1/2 ` `T_(2) = sqrt((T_1)/(4)) = 273/4 = 68.25 A` `=(68.25 - 273)^(@)C = -(204.75)^(@)C`. |
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| 45. |
Distinguish between average speed and rms speed. If three molecules have speed `u_(1), u_(2), u_(3)` what will be their average speed and rms speed. |
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Answer» Average speed is the arithmetic mean of the speeds of the molecules. `:.` Average speed = `(u_(1)+u_(2)+u_(3))/(3)` rms speed is the root mean square and is define as the square root of the mean of the wquares of different speeds of the individual molecules. rms speed = `sqrt((u_(1)^(2)+u_(2)^(2)+u_(3)^(2))/(3))`. |
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| 46. |
Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters `1 A and 2 A`. The gases may be considered under identical conditions of temperature, pressure and volume. |
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Answer» here, `d_(1) = 1 Å, d_(2) = 2 Å` under idential condition of temperature , pressure and volume, mean free path , `lambda prop 1/(d^2)` `:. (lambda_1)/(lambda_2) = (d_(1)^(2))/(d_(2)^(2)) = (2/1)^(2) = 4:1`. |
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| 47. |
A litre of dry air weighs 1.293 gram at S.T.P Find the temperature at which a litre of air will weigh one gram when the pressure is 72 cm. of mercury. |
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Answer» Correct Answer - `61.4^(@)C` At STP, volume of 1 gram of dry air, `V= (1000//1.293) c. C. As, (PV)/(T) = (P_(1)V_(1))/(T_1)` `:. T_(1) = (P_(1)V_(1)T)/(PV) = (72 xx 1000 xx 273)/(76 xx (1000//1.293))` `=334.4 K = 334.4-273 = 61.4^(@)C` |
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| 48. |
At `10^(@)C`, the value of the density of a fixed mass of an ideal gas divided by its pressure is x. at `110^(@)C`, this ratio isA. xB. `(383)/(283)x`C. `(10)/(110)x`D. `(283)/(383)x` |
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Answer» Correct Answer - D For a given mass m of an ideal gas of gas constant R, and density `rho` `(PV)/(T) = R or (Pm)/(rho T) = R or (ro)/(P) =(m)/(TR)` or, `x = (m)/(TR) or xT = m/R =a constant` `:. X_(1)T_(1) = x_(2)T_(2)` or ` x_(2) = x_(1) (T_1)/(T_2) = x ((273+10))/((273+110)) = (283x)/(383)`. |
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| 49. |
The pressure exerted by an ideal gas is numerically equal to ……………of…………………..per unit volume of the gas |
| Answer» Two third , average kinetic energy of translation | |
| 50. |
Assertion : Internal energy of an ideal gas does not depend upon volume of the gas. Reason : This is because internal energy of ideal gas depends only on temperature of gas.A. If, both Assertion and reason are ture and the Reason is the correct explanation of the Assertion.B. If both, Assertion and reason are true but Reason is not a correct expationation of the Assertion.C. If Assertion is true but the Reason is false.D. If Both, Assertion and reason are false. |
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Answer» Correct Answer - B Both, the assertion and reason are true. But the reason is not a correct explanation of the assertion. |
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