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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.
| 401. |
Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in the P-T diagram. During the process AB , pressure and temperature of the gas very such that `PT=Constant`. It `T_1=300K`, calculate (a) the work done on the gas in the process AB and (b) the heat absorbed or released by the gas in each of the processes. Give answer in terms of the gas constant R. |
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Answer» For the process A - B, it is given that PT = constant Differentiating above equation partially, we have PdT + TdP = 0 Equation of state for two moles of a gas `PV = 2 RT` or `P = (2RT)/(V)` After differentiating Eq. (ii) partially, we get PdV + VdP = 2R dT From Eq. (ii) partially, we get PdV + VdP = 2R dT From Eqs. (i) and (ii), we have `((2RT)/(V)) dT + T dP = 0` or 2RT dT + VTdP = 0 VdP = - 2 RdT Now form Eqs. (iii) and (iv), we have `- 2RdT + VdP = 2RdT` or `PdV = 4 RdT` a. The work done in the process AB `W_(AB) int PdV = int_(600)^(300) 4 RdT` `= 4 R |T|_(600)^(300) = 4 R (300 - 600)` `= - 1200 R` b. As process `B rarr C` is isobaric, so `Q_(BC) = nC_(P) Delta T = 2 xx (5R)/(2) xx (600 - 300)` = 1500 R ii. Process `C rarr A` is isothermal, so `Delta U = 0` `Q_(CA) = Delta U + W_(CA) = W-(CA)` `W_(CA) = nRT 1n (P_(C ) // P_(A))` `2R xx 600 1n (2P_(1) // P_(1)) = 1200 R1n 2` `Q_(CA) = 1200 R 1n 2` Again for process `A rarr B` `Q_(AB) = Delta U + W_(AB)` `= nC_(V) Delta T + W_(AB)` `= 2 xx ((3R)/(2)) xx (300 - 600) - 1200 R` `= - 900 R - 1200 R = - 2100 R` |
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| 402. |
A sample of an ideal monoatomic gas is taken round the cycle ABCA as shown in the figure the work done during the cycle is A. ZeroB. 3PVC. 6PVD. 9PV |
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Answer» Correct Answer - B `DeltaW=` Area of the loop `=(1)/(2)ACxxBC` |
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| 403. |
The work done by a gas taken through the closed process ABCA, see figure is A. `6P_(0)V_(0)`B. `4P_(0)V_(0)`C. `P_(0)V_(0)`D. zero |
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Answer» Correct Answer - A `W=(1)/(2) xx3V_(0)xx4P_(0)=6P_(0)V_(0)` |
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| 404. |
Consider a vertical tube open at both ends. The tube consistss of two parts, each of different cross sections and each part having a piston which can move smoothly in respective tubes. The two piston which can move smoothly in respective tube wire. The piston are joined together by an inextensible wire. The combined mass of the two piston is `5 kg`and area of cross section of the upper piston is `10 cm^(2)` greater than that of the lower piston. Amount of gas enclosed by the pistons is `1 mol`. When the gas is heated slowly, pistons move by `50 cm`. Find the rise in the temperature of the gas in the form `X//R K`, where `R` is universal gas constant. Use `g = 10 m//s^(2)` and outside pressure `= 10^(5) N//m^(2))`. |
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Answer» Because the heating pressure inside is not changed, let inside pressure be `rho`. Then for equilibrium of the systme. `P (A_(1) - A_(2)) = P_(0) (A_(1) - A_(2)) + (m_(1) + m_(2)) g` `P Delta V = (rho_(0) Delta A + mg) l` `l` is the displacement of the piston. `P Delta V = nR Delta T` `Delta T = (P Delta V)/(n R) = ((rho_(0) Delta A + mg) l)/(n R)` `= ((10^(5) Pa xx 10^(-3) m^(2) + 5 xx 10) (50 xx 10^(-2)))/(1 xx R)` `Delta T = (75)/(R ) K` |
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| 405. |
Gas at a pressure `P_(0)` in contained as a vessel. If the masses of all the molecules are halved and their speeds are doubles. The resulting pressure P will be equal toA. `4 P_(0)`B. `2 P_(0)`C. `P_(0)`D. `P_(0)//2` |
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Answer» Correct Answer - B `P=1/3 (mN)/(V) v_(rms)^(2) :. P prop mv_(rms)^(2)` so `(P_2)/(P_1) = (m_2)/(m_1) xx ((v_2)/(v_1))^(2) = (m_(1)//2)/(m_1)((2v_(1))/(v_1))^(2) = 2` `implies P_(2) = 2P_(1) = 2P_(0)`. |
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| 406. |
The density of an ideal gas is `1.25xx10^(-3)g cm^(-3)` at STP. Calculate the molecular weight of the gas. |
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Answer» From the formula ` PV= (nRT) = (m/M)RT ……..(1)` where m= mass of the gas M= molecular weight of the gas From equation (1), ` P=(m/M) ((RT)/V)= ((dRT)/M)` ` :. M= (dRT/P)= ((1.25 xx 10^3 xx 8.3 xx 273)/10^5)` `= 28.3 g/mol` |
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| 407. |
Consider `P-V` diagram for an ideal gas shown in figure. Out of following diagrams(figure). Which represents the `T-P` diagram? A. B. C. D. |
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Answer» Correct Answer - C Accoring to the question given that pV-constant Hence, we can say that the gas is going through an isothermal process. Clearly, form the graph that between process 1 and 2 temperature is constant and the gas expands and pressure decreases i.e., `P_(2)ltP_(1)` which corresponds to diagram (iii). |
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| 408. |
An ideal gas underoges cyclic process of ABCDA as shown in Given `P-V` diagram (figure) The amount of work done by the gas isA. 6PoVoB. `-2PoVo`C. `+2 PoVo`D. `+4 PoVo` |
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Answer» Correct Answer - B Consider the p-V diagram given in the question. Work done in the process ABCD=-area of reactangle ABCDA. `(AB)xxBC=(3V_(0)-V_(0))xx(2P_(0)-P_(0))` `=2V_(0)xxP_(0)=2P_(0)V_(0)` As the process is going anit -clockwise, hence there is a net compression in the gas, So, work done by the gas `=-2P_(0)V_(0)` |
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| 409. |
Refer to the plot of temperature versus time (figure) showing the changes in the state if ice on heating (not to scale). Which of the following is correct ? .A. The region AB represent ice and water in thermal equilibriumB. At B water starts boilingC. At C all the water gets converted into steamD. C to D represents water and steam in equilibrium at biling point. |
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Answer» Correct Answer - A::D During the process AB temperature of the system is `0^(@)C` Hence, it represetns phase change that is transformation of ice into water while temperature remians `0^(@)C`. BC represents rise in temperature of water from `0^(@)C` to `100^(@)C`(at C) Now, water starts converting into steam which is represetn by CD. |
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| 410. |
The density of oxygen at N.T.P. is `16xx0.089kg//m^(3)` then the R.M.S. velocity of oxygen molecules at N.T.P. would beA. `4.6xx10^(2)m//s`B. `46xx10^(2)m//s`C. `0.46xx10^(2)m//s`D. `3.6xx10^(2)m//s` |
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Answer» Correct Answer - A `C=sqrt((3P)/(rho))=sqrt((3xx10^(5))/(16xx90xx10^(-3)))` `C=sqrt((3xx10^(7))/(16xx93))=sqrt((10xx10^6)/(16xx3))=sqrt((3.33)/(16)xx10^(6))` `=(1.8xx10^(3))/(4)=0.45xx10^(3)=450m//s`. |
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| 411. |
if the density of airr at N.T.P. is 1.25`kg//m^(3)`, then the r.m.s. velocity of air molecules at N.T.P. will beA. 0.50 km/sB. 0.48 km/sC. 0.96 km/sD. 0.64 km/s |
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Answer» Correct Answer - B `C=sqrt((3P)/(rho))=sqrt(3xx1.013xx10^(5))/(1.25))=0.4847km//s` |
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| 412. |
What will be the r.m.s. speed of a gas at `800^(@)K` ?A. four times the values at 200 KB. half the value at 200 KC. twice the value at 200 KD. same as at 200 K |
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Answer» Correct Answer - C `C_(1)=K sqrt(800) and C_(2) =K sqrt(200)` `(C_(2))/(C_(1))=sqrt((200)/(800))=(1)/(2) C_(1)=2C_(2)` |
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| 413. |
One of the two vessels of same capacity is filled with oxygen and other is filled with helium of same mass. Both gasess are at same temperature. The ratio of pressure of these gases isA. `(P_(0))/(P_(He))=1/8`B. `(P_(0))/(P_(He))=1/4`C. `(P_(0))/(P_(He))=4/1`D. `(P_(0))/(P_(He))=8/1` |
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Answer» Correct Answer - A `(P_(0))/(P_(He))=(M_(He))/(M_(O_(2)))=4/32=1/8` |
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| 414. |
Two vessels have equal volums. One of them contains hydrogen at one atmosphere and the other helium at two atmosphere. If both the samples are at the same temperature, the `rms` velocity of the hydrogen molecules isA. equal to that of heliumB. twice that of heliumC. half that of heliumD. `sqrt(2)` times that of helium |
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Answer» Correct Answer - D `(V_(1))/(V_(2))=sqrt((M_(2))/(M_(1)))=sqrt((4)/(2))=sqrt(2)V_(1)=sqrt(2)V_(2)` |
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| 415. |
Which one of the following is not an assumption in the kinetic theory of gases?A. The volume occupied by the mplecules is negligibleB. The force of attraction between the melecules is negligibleC. The collision between molecules are elasticD. All molecules have some speed |
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Answer» Correct Answer - D All molecules have some speed . This statement is true but this is not an assumptionof kinetic theory . |
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| 416. |
One kg of a diatomic gas is at pressure of `8xx10^4N//m^2`. The density of the gas is `4kg//m^3`. What is the energy of the gas due to its thermal motion?A. `3xx10^(4)J`B. `5xx10^(4)J`C. `6xx10^(4)J`D. `7xx10^(4)J` |
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Answer» Correct Answer - B Thermal energy correspods to internal energy Mass=1kg, Density `=4kgm^(-3)` `rArr" " "Volume"=("mass")/("density")=(1)/(4)m^(3)` pressure=`8xx10^(4)Nm^(-2)` `therefore` Internal energy `=(5)/(2)PxxV=(5)/(2)xx(8xx10^(4))/(4)=50+10^(4)"Joule"` |
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| 417. |
One kg of a diatomic gas is at pressure of `8xx10^4N//m^2`. The density of the gas is `4kg//m^3`. What is the energy of the gas due to its thermal motion? |
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Answer» Energy of diatomic gas due to its thermal motion is `=(5)/(2)PVrArr(5)/(2)((m)/(rho))=5xx10^(4)J` |
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| 418. |
A 60 w bulb has a filament temperature of 2000 K. what will be the wattage of another bulbs of same filament area and material at a temperature of 4000 K?A. 8 wB. 32 wC. 64 wD. 960 w |
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Answer» Correct Answer - D `(P_(2))/(P_(1))=(T_(2)^(4))/(T_(1)^(4))=((4000)/(2000))^(4)=16` `P_(2)=16xxP_(1)=16xx60=960w` `P=sigma A rhoT^(4)` |
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| 419. |
An ideal gas is expanding such that `PT^(2)= `constant. The coefficient of volume expansion of lthe gas is:A. `(1)/(T)`B. `(2)/(T)`C. `(3)/(T)`D. `(4)/(T)` |
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Answer» Correct Answer - C |
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| 420. |
At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at `47^(@)C`A. 20KB. 80KC. `-73K`D. `3K` |
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Answer» Correct Answer - A |
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| 421. |
At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at `47^(@)C`?A. `20 K`B. `80 K`C. `-73 k`D. `3 k` |
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Answer» Correct Answer - A For oxygen `V_(O_2) = sqrt((3RT_(O_2))/(M_(O_2)))` and For hydrogen, `v_(H_2) = sqrt(3R(T_(H_2))/(M_(H_2)))` According to problem `= sqrt((3RT_(O_2))/(M_(O_2))) = sqrt(3R(T_(H_2))/(M_(H_2)))` `implies (T_(O_2))/(M_(O_2)) = (T_(H_2))/(M_(H_2)) implies (47+273)/(32) = (T_(H_2))/(M_(H_2))` `implies T_(H_2) = (320)/(32)xx2 = 20K`. |
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| 422. |
At what temperature is the root mean square velocity of gaseous hydrogen molecules is equal to that of oxygen molecules at `47^(@)C`?A. `20 k`B. `80 K`C. `- 73 K`D. `3 K` |
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Answer» Correct Answer - A `(v_(rms))_(H_(2)) = (v_(rms))_(O_(2))` `sqrt((3RT_(H_(2)))/((M_(0))_(H_(2)))) = sqrt((3 R T_(O_(2)))/((M_(0))_(O_(2))))` `sqrt(T_(H_(2))/(2)) = sqrt(((273 + 47))/(32))` `T_(H_(2)) = 20 K` |
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| 423. |
Calculate the rms velocity of molecules of a gas of density `1.5 g litre^(-1)` at a pressure of `2 xx 10^(6) N//m^(2)`.A. `2xx10^(2)m/s`B. `2xx10^(2)cm//s`C. `2xx10^(3)m//s`D. `2xx10^(3)cm//s` |
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Answer» Correct Answer - C `v_(rms)=sqrt((3P)/(rho))` |
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| 424. |
A 10kw drilling machine is used for 5 minutes to bore a hole in an aluminium block of mass `10xx10^(3)kg`. If 40% of the work done is utilised to raise the temperature of the block, then find the raise in temperature of the aluminium block? (Specific heat of Aluminium `=0.9 Jkg^(-1)k^(-1)`) |
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Answer» Work done by the drilling machine in 5 min W=power x time `=10xx10^(3)xx5xx60=3xx10^(6)J` The energy utilised to rise the temperature of the block `=40% "of" W=3xx10^(6)xx(40)/(100)=12xx10^(5)J` Heat gained by aluminimum block= mass xx specific heat xx increase in temperature `12xx10^(5)=(10xx10^(3))xx0.9xxDeltat` `therefore Deltat=(12xx10^(5))/(0.9xx10^(4))=133.3^(@)C` |
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| 425. |
The temperature and the relative humidity of air are `20^(@)c and 80%` on a certain day. Find the fraction of the mass of water vapour that will condence if the tempreture falls to `5^(@)C are 17.5mm`and 6.5mm of mercury respectively. |
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Answer» The relative humidity is (vapour pressure of the air)/(SVPat the same temperature) Thus, the vapour pressure at `20^(@)C` `=0.8xx17.5mmof Hg` `=14mmof Hg.` Consider a volume V of air, If the vapour pressure is p and the tempreture is T, the mass m of the vapour present is giving by `pV=(m)/(M)RT` or, `m=(MV)/(R) (p)/(T). ...(i)` the mass present at 20^(@)C i `m_(1)=(MV)/(R) (14mm of Hg)/293K)` When the air is cooled to `5^(@)C,` some vapour condenses and the air gets saturated with the vapour pressure at 5^(@)C is, therefore, 6.5mm of mercury. The mass pf vapour present at 5^(@)C is, therefore, `m_(2)=(MV)/(R) (6.5mm of Hg)/(278K).` THe fraction condensed `=(m_(1)-m_(2))/(m_(1))=1-(m_(2))/m_(1))` `=1-(6.5)/(278)xx(293)/(14)=0.51. |
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| 426. |
A vessel of volume `V = 7.51` contains a mixture of ideal gases at a temperature `T = 300 K : v_1 = 0.10` mole of oxygen, `v_2 = 0.20` mole of nitrogen, and `v_3 = 0.30` mole of carbon dioxide. Assuming the gases to be ideal, find : (a) the pressure of the mixture , (b) the mean molar mass `M` of the given mixture which enters its equation of state `p V = (m//M) RT`, where `m` is the mass of the mixture. |
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Answer» Correct Answer - (a)2.32 (b)1.7r (a) Let `p_(1),P_(2) and P_(3)` be the partical pressures of the 3 gases Then `P_(1) v= n_(1)RT,P_(2) = n_(2) RT and P_(3) V=n_(3)RT` `therefore P_(mix) = P_(1)+P_(2)+P_(3)=(n_(1)+n_(2)+n_(3))(RT)/(V)` `=(0.1+0.2+0.3)xx(8.3xx300)/(7.5xx10^(-3)) = 1.992xx10^5Nm^(-2) = 2 atm` (b) `M_(mix) = (n_(1)M_(1)+n_(2)M_(2)+n_(3)M_(3))/(n_(1)+n_(2)+n_(3))=(0.1xx32xx0.2xx28xx0.3xx44)/(0.1+0.2+0.3)=36.7` |
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| 427. |
Consider the lung capacity to be `500 cm_(3)` and the pressure to ve equivalent of 761mm of HG, estimate the number of molecules per breath. |
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Answer» From ideal gas equation, `PV=NkappaT` Normal body temperature is `98.6^(0)F=37^(0)C=310K.` `N=(PV)/(kT)` `=((101.46)(500xx10^(-6)))/((1.3807xx10^(-23)xx310))` `=1.19xx10^(19)`molecules |
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| 428. |
A steel tank contains 300 g of ammonia gas `(NH_(3))` at a pressure of `1.35 xx 10^(6)` Pa and a temperature of `77^(@)C` . Later the temperature is `22^(@)C` and the pressure is `8.7 xx 10^(5)` Pa .A. the initial volume of the tank is 38 litreB. finally 71 g grams of gas will leak out of the tank .C. the number of moles of the remaining gas is `13.5` mol .D. the number of moles of the remaining gas is `8.5` mol. |
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Answer» Correct Answer - A::B::C |
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| 429. |
A vessel of volume V = 30 litre is separated into three equal parts by stationary semi permeable membrane . The left , middle and right parts are filled with `m_(H) = 30` gm of hydrogen , `m_(O_(2)) = 160` gm of oxygen and `m_(N) = 70` gm of nitrogen respectively . The left partition lets through only hydrogen while the right partition lets through hydrogen and nitrogen . If the temperature in all is 300K , the ratio of pressure in the three compartments will be :A. `4 : 9: 5`B. `1.3 : 4.5 : 2`C. `9 : 4 : 5`D. `9: 5: 4` |
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Answer» Correct Answer - B |
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| 430. |
The initial temperature of a gas is `100^(@)C`. The gas is contained in closed vessel. If the pressure on the gas is increased by `5%` calculate the increase in temperature of the gasA. `1^(@)C`B. `2^(@)C`C. `4^(@)C`D. `5^(@)C` |
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Answer» Correct Answer - D Volume of gas is constant `T_(1)//T_(2) = P_(1)//P_(2)` `(100)/((100-DeltaT)) = (P)/(1.05P)` `105P =100P+P*Delta T` `(5P)/(P)=Delta T = 5^(@)C`. |
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| 431. |
The initial temperature of a gas is `100^(@)C` . The gas is contained in closed vessel . If the pressure of the gas is increased by `5%` , calculate the increase in temperature of the gas :A. `1^(@)C`B. `2^(@)C`C. `4^(@)C`D. `5^(@)C` |
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Answer» Correct Answer - D |
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| 432. |
Keeping the number of moles, volume and temperature the same, which of the following are the same for all ideal gases?A. `rms` speed of a moleculeB. densityC. pressureD. average magnitude of momentum |
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Answer» Correct Answer - C `PV = nRT` |
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| 433. |
According to the kinetic theory of gases, at absolute temperatureA. water freezesB. liquid helium freezesC. molecular motion stopsD. liquid hydrogen freezes |
| Answer» Correct Answer - C | |
| 434. |
Heat isA. Kinetic energy of moleculesB. potential and kinetic enrgy of moleculesC. energy in transtisD. work done on the system. |
| Answer» Correct Answer - C | |
| 435. |
Which of the following statements are true regarding the kinetic theory of gases?A. The pressure of the gas is directly proportional to the average speed of the moleculesB. The root mean square speed of the molecules directly proportional to the pressureC. The rate of diffusion is directly proportional average speed of the moleculesD. The average kinetic energy per molecule is inverse proportional to the absolute temperature. |
| Answer» Correct Answer - C | |
| 436. |
Of the following specific heat is maximum forA. mercuryB. CopperC. WaterD. Silver |
| Answer» Correct Answer - C | |
| 437. |
At a constant pressure of `10^(4)Nm^(-2)` , a gas expands by `0.25m^(3)` work done by the gas isA. 2500 JB. 250 J/kg KC. 25 JD. 2.5 J |
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Answer» Correct Answer - A `W=PdV` `=10^(4)xx0.25` =2500 J. |
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| 438. |
A L shaped container has dimensions shown in the fig. It has circular cross section of radius R. It is placed on a smooth horizontal table. The container is divided into two equal sections by a membrane AB. One section contains nitrogen and the other one contains oxygen. Temperature of both sides is same but pressure in the compartment having `N_(2)` is 4 times that in the other compartment. Due to some reason the membrane gets punctured. Find the distance moved by the cylinder if the cylinder is constrained to move in x direction only [with the help of guiding walls `W_(1) " and " W_(2)`]. There is no friction anywhere and mass of the cylinder and membrane is negligible |
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Answer» Correct Answer - 0.83 R towards left |
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| 439. |
The tempareture and humdity of air are `27^(@)C` and 50% on a particular day. Calculate the amount of vapour that should be added to 1 cubic metre of air to saturate.it. The saturation vapour present in the room. |
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Answer» ` R_H = VP/SVP` ` Given, 0.50 = VP/3600 ` `rArr VP = 3600 xx 0.50` `Let the extra pressure needed be P` So, ` P = (m/M) xx (RT/V) ` ` = (m/18) xx (8.3 xx 300 / 1)` Now, ` m/M xx 8.3 xx 300 xx 3600 xx 0.50 ` = 3600 ` [ air is saturated i.e. RH = 100% = 1` or VP = SVP] ` rArr m= ((36-18)6/8.3) ` ` = (18 xx 6/8.3) = 13 gm ` . |
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| 440. |
find the rms speed of oxygen molecules in a gas at 300K. |
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Answer» `v_(rms)=sqrt((3RT)/(m_(0)))` `sqrt((3xx(8.3JK^(-1)mol^(-1)xx(300K)))/(32g mol^(-1)` `sqrt((3xx8.3xx300)/0.032)ms^(-1)=483ms^(-1)` |
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| 441. |
A very tall vertical cylinder is filled with a gas of molar mass M under isothermal conditions temperature T. the density and pressure of the gas at the base of the container is `rho_(0)` and `P_(0)`, respectively Select the incorrect statement if gravity is assumed to be constant throughout the containerA. Both pressure and density decreases exponetially with heightB. The variation of pressure is `P=P_(0)e^(-(Mgh)/(RT))`C. The variation of density `rho=rho_(0)e^((Mgh)/(RT))`D. The molecular density decreases as one moves upwards. |
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Answer» Correct Answer - C Considering the equilibrium of differential layer of a thickness dh and height h, we have `(P+dP)A+dmg=PA` or `A dP=-g dm=0grhoA dh` From ideal gas equation, we have `P=(rhoRT)/(M)` or `rho=(PM)/(RT)` From eqns. (1) and (2), we get `dp=-(PM)/(RT)g dh` or `In (P)/(P_(0))=-(Mg)/(RT)h` or `P=P_(0)e^(-Mgh//RT)` In a similar manner if we eliminate P with the of ideal gas equation, we get `dP=(drhoRT)/(M)` The eqn. becomes `drho(RT)/(M)=-rhog dh` or `int_(P_(0))^(rho) (drho)/(rho)=-(Mg)/(RT)int_(0)^(h)dh` or `rho=rho_(0)e^(-Mgh//RT)` |
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| 442. |
A very tall vertical cylinder is filled with a gas of molar mass M under isothermal conditions temperature T. the density and pressure of the gas at the base of the container is `rho_(0)` and `P_(0)`, respectively Select the correct statementA. The density of gas cannot be uniform throughout the cylinderB. The density of gas cannot be uniform throughout the cylinder under isothermal conditionsC. The rate of change of density `|(drho)/(dh)|=(rhoMg)/(RT)`D. All of the above |
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Answer» Correct Answer - D Considering the equilibrium of differential layer of a thickness dh and height h, we have `(P+dP)A+dmg=PA` or `A dP=-g dm=0grhoA dh` From ideal gas equation, we have `P=(rhoRT)/(M)` or `rho=(PM)/(RT)` From eqns. (1) and (2), we get `dp=-(PM)/(RT)g dh` or `In (P)/(P_(0))=-(Mg)/(RT)h` or `P=P_(0)e^(-Mgh//RT)` In a similar manner if we eliminate P with the of ideal gas equation, we get `dP=(drhoRT)/(M)` The eqn. becomes `drho(RT)/(M)=-rhog dh` or `int_(P_(0))^(rho) (drho)/(rho)=-(Mg)/(RT)int_(0)^(h)dh` or `rho=rho_(0)e^(-Mgh//RT)` |
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| 443. |
The average kinetic energy of a gas `-23^(@)C` and 75Cm pressure is `5xx10^(-14)` erg for `H_(2)`. The mean kinetic energy of the `O_(2)` at `227^(@)C` and 150cm pressure will beA. `80xx10^(-14)eg`B. `20xx10^(-14)erg`C. `40xx10^(-14)erg`D. `10xx10^(-14)erg` |
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Answer» Correct Answer - D `E=(3)/(2)kTrArrEpropT` |
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| 444. |
A vessel of volume `2000cm^(3)` contains 0.1 mol of oxygen and 0.2mol of caron dioxide. If the temperature if the mixture is 300K, find its pressure. |
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Answer» We have `p=(nRT)/(V)`. The pressure due to oxygen is `p_(1)=((0.1mol)(8.3JK^(-1) mol^(-1)(300K)))/((2000xx10^(-6)m^(-3))=1.25xx10^(5) Pa. Similarly, the pressure in the vessel is `p=p_(1)+p_(2)` `=(1.25+2.50)xx10^(5)Pa=3.75xx10^(5)Pa.` |
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| 445. |
Air is filled in a bottle and it is corked at `35^(@)C`. If the cork can come out at 3 atmospheric pressure, then upto what temperature should the bottle be heated to remove the cork ?A. `325^(@)C`B. `851^(@)C`C. `651^(@)C`D. none of the these |
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Answer» Correct Answer - C `P_(1)V = nRT_(1) rArr 1 xx V = nR(35 + 273) = nR xx 308` (i) `P_(2)V = nRT_(2) rArr 3 xx V = nRT_(2)` (ii) `(ii)//(i)` `3 = T_(2)//308 rArr T_(2) = 914 K` `T_(2) = 924 - 273 = 651^(@)C` |
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| 446. |
An air bubble of radius 2.0mm is formed at the bottom of a 3.3 deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure `=1.0xx10^(5)Pa` and desnity of water`=1000kg m^(-3). |
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Answer» Given, P_1 = (10^5) + rho g h = (10^5)+ 1000` ` = 1.33 xx (10^5)Pa` ` (P_2) = (10^5)Pa` ` T_1 = (T_2) = T ` ` V_1 = (4/ (3 pi))((2 xx (10^-3))^3), ` ` V_2 (4/(3pi(r^3)))` ` We know, ((P_1)(V_1)/T_1) = ((P_2)(V_2)/T_2)` ` ( 1.33 xx (10^5) xx (4/ 3pi) xx ((2 xx (10^-3))^3))/T_1` ` = ((10^5) xx (4/3 pi (r^3))/ T_2)` ` rArr = (1.33 xx 8 xx (10^5) xx (10^-9))` ` = (10^5) xx (r^3)` ` rArr r = (3 sqrt (10.64 xx (10^-9)))` ` = (2.19 xx (10^-3)) = 2.2 mn.` |
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| 447. |
Two ballons are filled, one with pure He gas and other by air, repectively. If the pressure and temperature of these ballons are same then the number of molecules per unit volume is:A. more in the He filled balloonB. same in both balloonsC. more in air filled balloonD. in the rato `1 : 4` |
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Answer» Correct Answer - B `PV = NkT` `(N)/(V) = (P)/(kT)` |
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| 448. |
Two ballons are filled, one with pure He gas and other by air, repectively. If the pressure and temperature of these ballons are same then the number of molecules per unit volume is:A. more in the He filled balloonB. same in both balloonsC. more in air filled balloonsD. in the ratio of 1:4 |
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Answer» Correct Answer - B Ideal gas equation can be written as pv=nRT or `n/v=p/(RT)`=constant So, at constant pressure and temperature, all gases, will contain equal number of molecules per unit volume. |
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| 449. |
Two ballons are filled, one with pure He gas and other by air, repectively. If the pressure and temperature of these ballons are same then the number of molecules per unit volume is:A. more in the He filled balloonB. same in both ballonsC. more in air filled balloonD. in the ratio of 1:4 |
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Answer» Correct Answer - B Ideal gas equation can be written as pV=nRT ….. (i) In this equation , n= number of moles of the gas p= pressure of the gas V = volume of the gas R= universal gas constant and T = temperature of the gas from Eq . (i) , We have , `(n)/(V)=(P)/(RT)="constant"` So , at constant pressure and temperature , both balloons will contain equal number of molecules per unit volume. |
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| 450. |
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. `x`B. `(383)/(283) x`C. `(10)/(110) x`D. `(283)/(383) x` |
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Answer» Correct Answer - D Writing ideal 1 gas law, `PV = nRT` `or `PV = m/M RT or `(PV)/(m) = 1/mRT` or `p/(rho) = (RT)/(M) or (rho)/(P)= 1/T` `:. (rho_(1)//P_(1))/(rho_(2)//P_(2)) = (T_2)/(T_1)` `implies (x)/((rho_(2)//P_(2))) = (383)/(283) or (rho_2)/(P_2) = (283)/(383)x`. |
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