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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. |
To what temperature in `.^(@)C` a gas has to be heated to produce 10% change in volume at constant pressure, if the initloa temperature of the gas is at `0^(@)C`?A. 17.3B. 27.3C. 33.3D. 13.15 |
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Answer» Correct Answer - B `(V_(2))/(V_(1))=(T_(2))/(T_(1))` `(V_(2)-V_(1))/(V_(1))=(T_(2)-T_(1))/(T_(1))` `(10)/(100)=(T_(2)-T_(1))/(T_(1))` On solving `T_(2)=27.3^(@)C`. |
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| 2. |
If the pressure of an ideal gas is decreased by 10% isothermally, then its volume willA. decrease by 9%B. increase by 10%C. increase by 11.6%D. increase by 9% |
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Answer» Correct Answer - C `(dV)/(V)=(dP)/(P_(2))` `=(10)/(0.9)%=(100)/(9)=11.6%` |
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| 3. |
By what percentage should the pressure of a given mass of a gas be increased so as to decrease its volume by 10% at a constant temperature?A. `8.1%`B. `10.1%`C. `9.1%`D. `11.1%` |
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Answer» Correct Answer - D `(DeltaP)/(P)=(DeltaV)/(V_(2))=(1)/(0.9)=(10)/(9)=11.1%` |
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| 4. |
A certain volume of dry air at `20^(@)C` is expanded to three thimes its volume (i) Slowly, (ii) suddenly. Calculate the final pressure and temperature in each case. Atmosperic pressure `= 10^(5) Nm^(-2), gamma` of air = 1.4 |
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Answer» Correct Answer - `2.15xx10^(4N),^(-2)` When the process is slow, temperature reamins constant and it maty be taken as an isothermal process for which pV=a cosntant `therefore 10^(5)V=p_(2)3 V or p_(2)=(10^(5)/(3)=3.3xx10^(4)Nm^(-2)` When the process is sudden, it adiabatic In a adiabatic change ltbRgt `TV^(gamma-1)= a constant` or `293V^(0.4)=(273+t)3^(0.4),or(273+t)=(293)/(3)^(0.4)` `therefore log(273+t)=log293-log3^(0.4)=log293-0.4log3` `therefore 273+1 = antilog 2.2761 = 188.8` For an adiabatic change `pV^(gamma)=constant` `therefore 10^(5)V^(1.4)=p_(2)(3V)^(1.4)`, or `10^(5)V^(1.4)V^(1.4)` or `p_(2)=106(5)/(3)6(1.4)` `therefore logp_(2)-log 106(5)-1.4log3=5-1.4xx0.4771=5-0.6679=4.3321` `therefore P_(2) = antilog 4.3321 = 2.15xx10^(4)Nm^(-2)` In the slow process final temperature `= 20^(@)C` and pressure `= 3.3xx106(4)Nm^(-2)` In the sudden process final temperature`=-84.2^(@)C` and pressure `= 3.3 xx10^(4)Nm^(-2)` In the sudden process final tempeature`=-84.26^(@)C` and pressure `= 2.15xx10^(4)Nm^(-2)` |
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| 5. |
The volume of a gas at `20^(@)C` is 100 cm 3 at normal pressure. If it is heated to `100^(@)C` , its volume becomes 125 cm 3 at the same pressure, then volume coefficient of the gas at normal pressure isA. `0.0033//.^(@)C`B. 0.0030/`.^(@)C`C. 0.0025/`.^(@)C`D. 0.0021/`.^(@)C` |
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Answer» Correct Answer - A `v=(DeltaV)/(Vdeltat)=(25)/(100xx80)=0.0033//.^(@)C` |
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| 6. |
One litre of helium gas at a pressure `76 cm`. Of Hg and temperature `27^(@)C` is heated till its pressure and volume are double. The final temperature attained by the gas is:A. `327^(@)C`B. `927^(@)C`C. `1027^(@)C`D. `827^(@)C` |
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Answer» Correct Answer - B Initial volume of the gas `V_(1) = 1` litre initial pressure the gas `P_(1) = 70 "cm of Hg"` Final pressure of the gas `P_(2) = 2 xx 76 = 152 "cm of Hg"` Final volume of gas `V_(2) = 2 "litre"` According the Gay Lussac law we have `(P_(1)V_(1))/(T) = (p_(2)V_(2))/(T_2) or (76XX)/(300) = (152XX2)/(t_2)` `(1)/(300) = 4/(t_2) OR t_(2)B = 4 XX 300 = 1200k = 927^(@)c`. |
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| 7. |
An ideal gas has volume`V_(0)` at . `27^(@)C` It is heated at constant pressure so that its volume becomes . `2V_(0).`The final temperature isA. `327K`B. `327^(@)C`C. `54^(@)C`D. `150^(@)C` |
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Answer» Correct Answer - B `(V_(2))/(V_(1))=(T_(2))/(T_(1))` `T_(2)=(V_(2))/(V_(1))" "T_(1)=(2V_(o))/(V_(o))xx300` `T_(2)=600K` `therefore t_(2)=600-273=327^(@)C` |
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| 8. |
`250 litre` of an ideal gas is heated at constant pressure from `27^(@)C` such that its volume becomes `500 litre`. The final temperature isA. `54^(@)C`B. `300^(@)C`C. `327^(@)C`D. `600^(@)C` |
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Answer» Correct Answer - C `V` doubled, `T` doulbed `T_(1) = (27+273)K = 300K` `T_(2) = 2 xx 300K = 600K` `t_(2) = (600-273)^(@)C = 327^(@)C`. |
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| 9. |
A gas at pressure P is abiabatically compressed so that its density becomes twice that of initial value. Given that `gamma=C_(p)//C_(v)=(7//5)` , What will be the final pressure of the gas ?A. `2p`B. `(7)/(5)`C. 2.63 pD. p |
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Answer» Correct Answer - C For adiabatic process,`p_(2)V_(2)^(gamma)=p_(1)V_(1)^(gamma)` `therefore"final pressure " p_(2)=p_(1)((V_(1))/(V_(2)))^(gamma)=p_(1)(rho_(2)/(rho_(1)))^(gamma)` `p=((2)/(1))^(7//5)=2.63p` |
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| 10. |
To what temperature should the oxygem at `27^(@)C` be heated at constant pressure, so that the root mean square velocity of its molecules becomes thrice of its previous value.A. `2700^(@)C`B. `2700K`C. `2327^(@)C`D. 270K |
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Answer» Correct Answer - B `v_(rms)=sqrt((3RT)/(M))rArrv_(rms)propsqrt(T)` |
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| 11. |
Root mean square velocity of gas molecules is `300 m//sec`. The `r.m.s` velocity of molecules of gas with twice the molecular weight and half the absolute temperature is :A. `300ms^(-1)`B. `600ms^(-1)`C. `75ms^(-1)`D. `150ms^(-1)` |
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Answer» Correct Answer - D rms velocity `C=sqrt((3pV)/(M))=sqrt((3RT)/(M))` and the new rms velocity , `c_(1)=sqrt((3R(T//2))/((2M)))=(1)/(2)sqrt((3RT)/(M))` `=(c)/(2)=(300)/(2)=150ms^(-1)` |
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| 12. |
The temperature at which the root mean square velocity of the gas molecules would becomes twice of its value at `0^(@)C` isA. `819^(@)C`B. `1092^(@)C`C. `1100^(@)C`D. `1400^(@)C` |
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Answer» Correct Answer - A `(C_t)/(C_0) = sqrt((273+t)/(273))` or `4 xx 273 - 273 = t` or `t = 3 xx 273 = 819^(@)C`. |
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| 13. |
Three perfect gases at absolute temperature `T_(1), T_(2)` and `T_(3)` are mixed. The masses f molecules are `m_(1), m_(2)` and `m_(3)` and the number of molecules are `n_(1), n_(2)` and `n_(3)` respectively. Assuming no loss of energy, the final temperature of the mixture isA. `((T_(1) + T_(2) + T_(3)))/(3)`B. `(n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))/(n_(1) + n_(2) + n_(3))`C. `(n_(1)T_(1)^(2) + n_(2)T_(2)^(2) + n_(3)T_(3)^(2))/(n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))`D. `(n_(1)^(2)T_(1)^(2) + n_(2)^(2)T_(2)^(2) + n_(3)^(2)T_(3)^(2))/(n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))` |
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Answer» Correct Answer - B `n_(1).(3)/(2)kT_(1) + n_(2)(3)/(2)kT_(2) + n_(3).(3)/(2)kT_(3)` `= (n_(1) + n_(2) + n_(3)).(3)/(2)kT` `T = (n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))/(n_(1) + n_(2) + n_(3))` |
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| 14. |
Three perfect gases at absolute temperature `T_(1), T_(2)` and `T_(3)` are mixed. The masses f molecules are `m_(1), m_(2)` and `m_(3)` and the number of molecules are `n_(1), n_(2)` and `n_(3)` respectively. Assuming no loss of energy, the final temperature of the mixture isA. `(n_(1)T_(1)+n_(2)T_(2)+n_(3)T_(3))/(n_(1)+n_(2)+n_(3))`B. `(n_(1)T_(1)^(2)+n_(2)T_(2)^(2)+n_(3)T_(3)^(3))/(n_(1)T_(1)+n_(2)T_(2)+n_(3)T_(3))`C. `(n_(1)^(2)T_(1)^(2)+n_(2)^(2)T_(2)^(2)+n_(3)^(2)T_(3)^(2))/(n_(1)T_(1)+n_(2)T_(2)+n_(3)T_(3))`D. `(T_(1)+T_(2)+T_(3))/(3)` |
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Answer» Correct Answer - A we Have `(F)/(2)n_(1)kT_(1)+(F)/(2)n_(2)kT_(2)+(F)/(2)n_(3)kT_(3)` `(F)/(2)(n_(1)+n_(2)+n_(3))kT` The Final Temperature of the mixture `T=(n_(1)T_(1)+n_(2)T_(2)+n_(3)T_(3))` |
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| 15. |
A motor car tyre is pumped up to pressure of two atmosheres at `15^(@)C` when it suddenly bursts. Calculate the resulting drop in temperature of the escaping air `(gamma =14)`. |
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Answer» Correct Answer - 2.3734 As it bursts suddenly, the changes is adiabatic. We have `T^(gamma)p^(1-gamma) =` a constant `T_(1)^(gamma)P_(1)^(1-gamma) = T_(2)^(gamma)p_(2)^(1-gamma)` or `(273+15)^(gamma)(2P_(0))^(1-gamma) = T_(2)^(gamma)(2p_(0))^(1-gamma) Where p_(0) = 1 atmoshpere` or `288^(1.4) 2^(1-1.4)p_(0)^(1-gamma) = T_(2)^(1.4)p_(0)^(1-gamma)` `288^(1.4) 2^(-0.4) = T_(2)^(1.4)` or ` T_(2)^(1.4) = (288^(1.4))/(2^(0.4))` or `log T_(2) = log288 - 0.4log 2//1.41 =2.3734` |
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| 16. |
Two moles of helium are mixed with n moles of hydrogen. The root mean spure (rms) speed of the gas molecules in the mexture is `sqrt2` times the speed of sound in the mixture. Then value of n isA. 1B. `3//2`C. 2D. 3 |
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Answer» The root mean square speed of the gas `v" rms"=sqrt((3RT)/(M))` , `v" sound"=sqrt((gammaRT)/(M)),v" rms"=sqrt(2)v" sound"` Solving it , we get `sqrt(3)=sqrt(2gamma)` `rArr" " gamma=(3)/(2)` for the mixture. As, `gamma=(C_(p))/(C_(v))=(n_(1)C_(p_(1))+n_(2)Cp_(2))/(n_(1)+n_(2))xx(n_(1)+n_(2))/(n_(1)C_(v_(1)+n_(2))C_(v_(2)))` `gamma=(n_(1)C_(p_(1))+n_(2)Cp_(2))/(n_(1)C_(v_(1))+n_(2)C_(v_(2)))` For helium , `Cp_(1)=(5)/(2)R,C_(v_(1))=(3)/(2)R` For hydrogen, `Cp_(2)=(7)/(2)R,C_(v_(2))=(5)/(2)R` `therefore" " (3)/(2)=(2((5)/(2)R)+n((7)/(2)R))/(2((3)/(2)R)++n((7)/(2)R))=(10+7n)/(6+5n)` or 20+14n=18+15n `rArr` n=2 |
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| 17. |
A carnot engine has efficiency `1//5` . Efficiency becomes `1//3` when temperature of sink is decreased by 50 K What is the temperature of sink ?A. 325 KB. 375 KC. 300 KD. 350 K |
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Answer» Correct Answer - C The efficiency of Carnot engine , `eta=1-(T_(L))/(T_(H))` Where , `T_(L)` is temperature of sink and `T_(H)` is temperature of According to question , `(1)/(5)=1-(T_(L))/(T_(H))` …(i) `and " " (1)/(3)=1-(T_(L)-50)/(T_(H))` ....(ii) Form Eq . (i) , `T_(L)/T_(H)=(4)/(5)rArrT_(H)=(5)/(4)T_(L)` Substituting value of `T_(H)` in Eq .(ii) , we get `(1)/(3)=1-(T_(L)-50)/((5)/(4)T_(L))rArr(4(T_(L)-50))/(5T_(L))=(2)/(3)` `"or" " " T_(L)-50=(2)/(3)xx(5)/(4)T_(L)` `"or" " " T_(L)-(5)/(6)T_(L)=50` `therefore" " T_(L)=50xx6=300K ` |
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| 18. |
Two identical vessels contain the same gas at pressure `P_(1)` and `P_(2)` at absolute temperature `T_(1) and T_(2)`, respectively. On joining the vessels with a small tube as shown in the figure. The gas reaches a common temperature T and a common pressure P. Determine the ratio `P//T` A. `[(P_(1) T_(1) + P_(2) T_(2))/(T_(1) T_(2))]`B. `(1)/(2)[(P_(1) T_(1) + P_(2) T_(2))/(T_(1) T_(2))]`C. `(1)/(2)[(P_(1) T_(2) + P_(2) T_(1))/(T_(1) T_(2))]`D. `[(P_(1) T_(2) + P_(2) T_(1))/(T_(1) T_(2))]` |
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Answer» Correct Answer - C On combining the two vessels the total number of moes remains constant , `i.e., n = n_(1)+n_(2)` Using gas equation, we can write , `n_(1) +(P_(1)V)/(RT_(1)), n_(2) = (P_(2)V)/(RT_(2))` and `n=(P(2V))/(RT)` where `V` is the volume of each vessel. Thus `(P(2V))/(RT) = (P_(1)V)/(RT_1)+(P_(2)V)/(RT_2) or P/T = 1/2[(P_1)/(T_1)+(P_2)/(T_2)]` . |
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| 19. |
A system absorbs 100 cal of heat and does an external work of 150 J. if J=4.2 J/cal the change in internal energy isA. 420 JB. 270 JC. 250 JD. 150 J |
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Answer» Correct Answer - B `DeltaQ=100xx4.2` `=420J` `DeltaW=150J` `DeltaU=DeltaQ-DeltaW` `=420-150=270J`. |
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| 20. |
One mole of an ideal monoatomic gas requires 207 J heat to raise the temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K, the heat required is [Given the gas constant R = 8.3 J/ mol. K]A. 215.3 JB. 198.7 JC. 207 JD. None of these |
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Answer» Correct Answer - B As, `C_(V)=C_(p)-R=207-8.3=198.7 J` |
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| 21. |
One mole of an ideal monoatomic gas requires 207 J heat to raise the temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K, the heat required is [Given the gas constant R = 8.3 J/ mol. K]A. 96.6 JB. 124 JC. 198.8 JD. 215.4 J |
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Answer» Correct Answer - B `n=1,dQ_(P)=207J,DeltaT=10K` `dQ_(P)=dQ_(V)+dw` `dQ_(V)=dQ_(P)-dw` `=207-nRdT` `207-1xx8.3xx10` `=207-83` `=124J`. |
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| 22. |
When heat energy of 1500 J is supplied to a gas it is compressed from 10 `m^(3)` to 5 `m^(3)` at a pressure of 100 Pa. the change in internal energy isA. 2000 JB. 1000 JC. 500 JD. 2500 J |
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Answer» Correct Answer - A `PDeltaV=P(V_(2)-V_(1))` `=100(5-10)` =-500J `DeltaU=DeltaQ-DeltaW` =1500-(-500) =2000J. |
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| 23. |
A gas at constant pressure of `4.5xx10^(5)Pa` to a gas it is compressed from 10 `m^(3)` to `3.0m^3` on givinig a heat of 800 kJ. The change in internal energy isA. `5.25xx10^(5)J`B. `6.75xx10^(5)J`C. `1.25xx10^(5)J`D. `3.25xx10^(5)J` |
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Answer» Correct Answer - C `DeltaW=+PDeltaV` `=4.5xx10^(5)xx1.5` `=6.75xx10^(5)` `DeltaQ=800kJ` `=8xx10^(5)J` `DeltaU=8.0xx10^(5)-6.75xx10^(5)` `DeltaU=1.25xx10^(5)J`. |
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| 24. |
The specific heat of a gas constant volume is 20 J/mol K. when two moles of such gas is heated through `10^(@)C` at constant pressure, what is the increase in thernal energy and work done? (R=8 J/mol K)A. 200J, 160 JB. 400 J, 260 JC. 400 J, 160 JD. 200 J, 460 J |
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Answer» Correct Answer - C `du=nC_(V)dT` `=2xx20xx10=400`J `domega=nRdT` `=2xx8xx10=160J` |
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| 25. |
At what temperature is the R.M.S. speed of gas molecules half the value at S.T.P.?A. 68.25 KB. 273 KC. 345 KD. 0 K |
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Answer» Correct Answer - A `(C_(2))/(C_(1))=sqrt((T_(2))/(T_(1)))` `(1)/(2)=sqrt((T_(2))/(T_(1)))` `(1)/(4)=(T_(2))/(T_(1))` `T_(2)=(T_(1))/(4)` `=(273)/(4)=68.25K` |
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| 26. |
What is the true for 3 moles of a gas?A. `3(C_(p)-C_(V))=R`B. `((C_(p)-C_(V)))/(3)=R`C. `C_(p)-C_(V)=R`D. `C_(p)-3C_(V)=R` |
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Answer» Correct Answer - C `C_(P)-C_(V)=R` |
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| 27. |
A berometer correctlu reads the atmospheric pressure as 76cm of mercury. Water droplets are slowly introduced into the barometer tube by a dropper. The height of the mercury column first decreases and then become constant. If the saturation vapour pressure at the atmospheric tgemperature is 0.80cm of mercury, find the height if the mercury column when it reaches its minimum value. |
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Answer» Pressure is minimum when the vapour present inside is at saturation vapour pressure. As this is the maximum pressure which the vapour can exert. Hence the normal level of mercury drops down by 0.8 cm :. The height of Hg column (Given SVP at atmospheric temperature = 0.08 cm of Hg ) = 76 - 0.80 cm = 75.2 cm of Hg. |
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| 28. |
A barometer tube contains a mixture of air and saturated water vopour in the space above the mercury column. It reads `70cm` when the actual atmospheric pressure is `76cm` of mercury. The saturation vapour pressure at room temperature is `1.0 cm` of mercury, The tube is now lowered in the reservoir till the space above the mercury column is reduced to half its original volume. Find the reading of the barometer. Assume that the temperature remains constant. |
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Answer» "The pressure due to the `air + vapour` is `76cm-70cm= 6m` of mercury the vapour is saturated and the pressure due to it is `1cm`of mercury. The pressure due to the air is" "therefore, 5cm`of mercury". , "As the tube is lowered and the volume above the mercury is decreased, some of the vapour will condense. The remaining vopour will again exert a pressure of the volme which is halved. Thus" ,`P_(air)=2xxcm=10cm`"of merucry . The pressure due to the `air+vapout`reading is" `76cm-11cm=65cm` |
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| 29. |
A barometer tube contains a mixture of air and saturated water vopour in the space above the mercury column. It reads 70cm when the actual atmospheric pressure is 76cm of mercury. The saturation vapour pressure at roon tempature is1.0 cm of mercury, The tube is now lowered in the reservoir till the space above the mercury cloumn is reduced to half its origina volume. Find the reading of the barometer. Assume that the temperature remains constant. |
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Answer» The pressure due to the air +vapour is `76cm-70cm=6cm` of mercury, The vapour is saturted and the pressure due to it is 1cm of mercury. The pressure due to the air is, therefore, 5cm of mercury, As the tube is lowered and the volume above the mercury is decreased, Some of the vapour will condense. The remaining vapour will again exert a pressure of 1 cm of mercury . the pressure due to air is doubled as the volume is halved. Thus, pair`=2xx5cm=10cm` of mercury. The pressure due to air=vapour `=10cm+1cm=11cm`mercury. The barometer reading is `76cm-11cm=65cm.` |
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| 30. |
A barometer reads 75cm if mercury. When `2.0cm^(3)` of air at atmospheric pressure is introduced into space above the mercury level, the volume of this space cecones `50 cm^(3)`. find the length by which the mercury column desecds. |
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Answer» Let the pressure of the air in the barometer be p. We have , `pxx50cm^(3)=(75cm of mercury)xx(2.0cm^(3))` `p=3.0cm` of murcury. The atmospheric pressure is equal to the pressure due to the mercury column plus the pressure due to the air inside, Thus, the mercury column descends by `3.0cm`. |
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| 31. |
How much should the pressure be increased in order to decrease the volume of a gas 5% at a constant temprature ?A. `5%`B. `5.26%`C. `10%`D. `4.26%` |
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Answer» Correct Answer - B New volume , `V_(1)=V-(5)/(100)V=(95V)/(100)` New pressure , `p_(1)=(pV)/(V_(1))=(pV)/((95//100)V)=(100)/(95)p` `therefore%"increase in pressure"=((p_(1)-P)/(p))xx100=(P_(1)/(p)-1)xx100` `((100)/(95)-1)xx100=5.26%` |
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| 32. |
n moles of an ideal gas undergo a process in which the temperature changes with volume as `T=kv^(2)`. The work done by the gas as the temperature changes from `7_(0)` to `47_(0)` isA. `3nRT_(0)`B. `((5)/(2))nRT_(0)`C. `((3)/(2))nRT_(0)`D. Zero |
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Answer» Correct Answer - C `T=kV^(2) therefore dT=2kVdV` or `dV=(dT)/(2kV), P=(nRT)/(V), W-int PdV` |
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| 33. |
A vessel contains 1 mole of `O_2` gas (relative molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (relative molar mass 4) at temperature 2T has a pressure ofA. `(P)//(8)`B. `P`C. `2P`D. `8P` |
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Answer» Correct Answer - C `PV = nRT " " rArr (P_1)/(P_2) =(n_(1)T_(1))/(n_(2)T_(2))` |
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| 34. |
If the temperature of a gas is increased from `0^(@)C` to `273^(@)C`, then the ratio of average kinetic energy of the gas molecules isA. `1:4`B. `4:1`C. `1:1`D. `2:1` |
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Answer» Correct Answer - D `KE_(1) prop T_(1) and KE_(2) prop T_(2)` `(KE_(2))/(KE_(1))=(T_(2))/(T_(1))=(273+273)/(273)=(2xx273)/(273)=(2)/(1)` |
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| 35. |
If the temperature of a gas is increased from `0^(@)` to `273^(@)C`, then the ratio of change in the average kinetic energy of the gas molecule to the original isA. `1:4`B. `4:1`C. `1:1`D. `2:1` |
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Answer» Correct Answer - C `KE_(1) prop T_(1) and KE_(2) prop T_(2)` `(KE_(2))/(KE_(1))=(T_(2))/(T_(1))=(273+273)/(273)=(2xx273)/(273)=(2)/(1)` `(KE_(2)-KE_(1))/(KE_(1))=1` |
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| 36. |
A jar has a mixture of hydrogen and oxygen in the ratio 4:5. then the ratio of mean kinetic energy of hydrogen and oxygen molecule isA. `1:16`B. `5:4`C. `4:5`D. `1:1` |
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Answer» Correct Answer - D `KE_(1) prop T_(1) and KE_(2) prop T` `(KE_(1))/(KE_(2))=(T)/(T)=1` |
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| 37. |
A container has a mixture of two gases, hydrogen and oxygen at rom temperature. Which keeping the temperature. Which one of the followingg statements is true? (If `c_(H) and c_(O)` are the root mean square velocities of hydrogen and oxygen molecules respectively)A. `c_(H) gt c_(O)`B. `c_(H) lt c_(O)`C. `c_(O)=4c_(H)`D. `c_(O)=16c_(H)` |
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Answer» Correct Answer - A `(C_(H))/(C_(O))=sqrt((M_(O))/(M_(H)))=sqrt((32)/(2))=sqrt(16)` `(C_(H))/(C_(O))=4" "thereforeC_(H) gt C_(O)`. |
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| 38. |
A jar has a mixture of hydrogen and oxygen gas in the ratio of 1 : 5. The ratio of mean kinetic energies of hydrogen and oxygen molecules isA. `1:16`B. `1:5`C. `1:4`D. `1:1` |
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Answer» Correct Answer - D K.E. per molecule is independent of nature of the gas. |
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| 39. |
A sample of hydrogen at temperature T, volume V and pressure P have speed v and in a sample of oxygen at temperature T, volume 2 V and pressure 3 P. the root mean square velocity of the oxygen molecule isA. `(v)/(4)`B. `sqrt(6)v`C. `sqrt(3)v`D. `sqrt(2)v` |
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Answer» Correct Answer - A `(C_(2))/(C_(1))=sqrt((M_(1))/(M_(2)))=sqrt((2)/(32))=sqrt((1)/(16))=(1)/(4)` `C_(2)=(C_(1))/(4)`. |
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| 40. |
It temperature of 5 moles of a gas is raised through `100^(@)C` at constant pressure, then external work done by the gas will be (R=8.32 J/mol K)A. 832 JB. 4160 JC. 41.6 JD. 15280 J |
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Answer» Correct Answer - B `Pdv=nRdT` `dw=5xx8.32xx100=4160J` |
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| 41. |
Two spheres of the same metal have radii in the_ratio 1 : 2 Their heat capacities are in what ratioA. `1:2`B. `1:4`C. `1:6`D. `1:8` |
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Answer» Correct Answer - D `(C_(1))/(C_(2))=(m_(1))/(m_(2))=((r_(1))/(r_(2)))^(3)=((1)/(2))^(3)=(1)/(8)` |
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| 42. |
If the temperature of a hot body is raised would increased byA. 0.16B. 0.125C. 0.04D. 0.09 |
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Answer» Correct Answer - A `(R_(1))/(R_(2))=((T_(2))/(T_(1)))^(4)=(1.04)^(2)=1+4xx0.04` `(R_(2)-R_(1))/(R_(1))=0.16=16%`. |
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| 43. |
The wavelength of maximum emitted energy of a body at 700 K is `4.08 mu m`. If the temperature of the body is raised to 1400 K , the wavelength of maximum emitted energy will beA. `1.02 mum`B. `16.32mum`C. `8.16mum`D. `2.04mum` |
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Answer» Correct Answer - D `lamda_(m_(1))T_(1)=lamda_(m_(2))T_(2)` `therefore lamda_(m_(2))=(lamda_(m_(1))T_(1))/(T_(2))=4.08xx(700)/(1400)=2.04`m |
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| 44. |
Two solid spheres of radii `R_(1)` and `R_(2)` are made of the same material and have similar surfaces. These are raised to the same temperature and then allowed to cool under identical conditions. The ratio of their initial rates of loss of heat areA. `R_(1)//R_(2)`B. `R_(2)//R_(1)`C. `(R_(1)^(2))/(R_(2)^(2))`D. `(R_(2)^(2))/(R_(1)^(2))` |
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Answer» Correct Answer - C `(R_(1))/(R_(2))=(A_(1))/(A_(2))=(4pir_(1)^(2))/(4pir^(2))=((r_(2))/(r_(1)))^(2)` |
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| 45. |
A black body at 200 K is found to exit maximum energy at a wavelength of `14mu m` . When its temperature is raised to 1000 K , the wavelength at which maximum energy is emitted isA. `14mum`B. `70muF`C. `2.8mum`D. `7mum` |
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Answer» Correct Answer - C `lamda_(m_(1))T_(1)=lamda_(m_(2))T_(2)` `therefore lamda_(m_(2))=(lamda_(m_(1))T_(1))/(T_(2))=(14xx200)/(1000)` `=2.8mum`. |
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| 46. |
The volume of a given mass of air at temperature `27^(@)C` is 100 c.c. if its temperature is raised to `57^(@)C` maintaining the pressure constant, thhen the increase in its volume isA. 100 c.c.B. 130 c.c.C. 10 c.c.D. 30 c.c. |
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Answer» Correct Answer - C `(V_(2))/(V_(1))=(T_(2))/(T_(1))` `(V_(2)-V_(1))/(V_(1))=(T_(2)-T_(1))/(T_(1))=(330-300)/(300)` `V_(2)-V_(1)=(30V_(1))/(300)=10` c.c. |
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| 47. |
The kinetic energy per kg of nitrogen molecules at 175^(@)C` is (Molecular weight of nitroegn 28 and R=3820 J/k mole K)A. `19.968xx10^(6)J`B. `1.783xx10^(5)J`C. `1.9968xx10^(5)J`D. `1.678xx10^(5)J` |
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Answer» Correct Answer - C `K.E_(2)//kg=(3)/(2)(RT)/(M)=(3)/(2)xx(8320xx448)/(28)` `=(3xx8320xx224)/(28)` `=3xx8320xx8=24xx8320=199680` |
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| 48. |
The pressure of a gas filled in a closed vessel increase by 0.4% when temperature is increased by `1^(@) C`. Find the initial temperature of the gas.A. `23^(@)C`B. `250^(@)C`C. `-23^(@)C`D. `300K` |
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Answer» Correct Answer - C `(DeltaT)/(T)=(DeltaP)/(P)` `=(0.4)/(100)=(4)/(1000)` `(1)/(T)=(1)/(250)` `T=250K` `t=250-273=-23^(@)C`. |
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| 49. |
A given amount of a gas is heated till the volume and pressure are each increased by 1%. Then the temperature increases byA. `0.5%`B. `1%`C. 0.02D. 0.04 |
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Answer» Correct Answer - C `(DeltaT)/T=((DeltaV)/(V)+(DeltaP)/(P))` `=((1)/(100)+(1)/(100))=(2)/(100)=2%` |
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| 50. |
The r.m.s. speed of the molecules of an enclosed gas in `C_(rms)`. If pressure is made four times keeping the temperature constant, the r.m.s. speed will beA. `2C_(rms)`B. `4C_(rms)`C. `C_(rms)`D. `(C_(rms))/(2)` |
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Answer» Correct Answer - C `C_(rms)` is indepenent of pressure. |
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