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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. |
Starting with the same initial conditions, an ideal gas expands from volume `V_1` to `V_2` in three different ways, the work done by the gas is `W_1` if the process is purely isothermal, `W_2` if purely isobaric and `W_3` if purely adiabatic, thenA. `W_2gtW_1gtW_3`B. `W_2gtW_3gtW_1`C. `W_1gtW_2gtW_3`D. `W_1gtW_3gtW_2` |
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Answer» Correct Answer - A Work done is equal to area under the curve on `PV` diagram. `(a)` is the correct option. |
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| 2. |
5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be `T_1,` the work done in the process isA. `9/8RT_1`B. `3/2RT_1`C. `15/8RT_1`D. `9/2RT_1` |
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Answer» Correct Answer - A Number of moles of He `=5.6//22.4=1//4` Now `T(5.6)^(gamma-1)=T_(2)(0.7)^(gamma-1)` `T_(1)=T_(2)((1)/(8))^(2//3) implies 4T_(1)=T_(2)` Work done `=-(nR[T_(2)-T_(1)])/(gamma-1)=((1)/(4)R[3T_(1)])/((2)/(3))=-(9)/(8)RT_(1)` |
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| 3. |
Calorie is defined as the amount of heat required to raise temperature of 1 g of water by `1^@C` and it is defined under which of the following conditions?A. From `14.5^@C` to `15.5^@C` at 760 mm of HgB. From `98.5^@C` to `99.5^@C` at 760 mm of HgC. From `13.5^@C` to `14.5^@C` at 76 mm of HgD. From `3.5^@C` to `4.5^@C` at 76 mm of Hg |
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Answer» Correct Answer - A One calorie is the heat required to raise the temperature of `1g` of water from `14.5^(@)C` to `15.5^(@)C` at `760mm` of `hg` . |
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| 4. |
In which of the following process, convection does not take place primarilyA. sea and land breezeB. boiling of waterC. heating air around a furnaceD. warning of glass of bulb due to filament |
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Answer» Correct Answer - D Heating of glass bulb through filament is through radiation. |
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| 5. |
Earth receives `1400 W//m^2` of solar power. If all the solar energy falling on a lens of area `0.2m^2` is focused on to a block of ice of mass 280 grams, the time taken to melt the ice will be….. Minutes. (`Latent heat of fusion of ice`=`3.3xx10^5J//kg`.) |
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Answer» Solar power received by earth `=1400W//m^(2)` Solar power received by `0.2m^(2)` area `=(1400W//m^(2))(0.2m^(2))=280W` Mass of ice `=280g=0.280Kg` Here rerquired to melt ice `=(0.280)(3.3xx10^(5))` `=9.24xx10^(4)J` If `t` is the time taken for the ice to melt, we will have `(280)t=9.24xx10^(4)J[:. P=(E)/(t)]` `t=(9.24xx10^(4))/(280)s=330s=5.5min` |
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| 6. |
The figure below shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). The temperature is increased continuously form 0 to 500K at a constant rate. Ignoring any volume change, the following statement (s) is (are) correct to a reasonable approximation. A. the rate at which heat is absorbed in the range 0-100K varies linearly with temperature T.B. heat absorbed in increasing the temperature from 0-100K is less than the heat required for increasing the temperature from 400-500K.C. there is no change in the rate of heat absorption in range 400-500K.D. the rate of heat absorption increases in the range 200-300K. |
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Answer» Correct Answer - A::B::C::D Option `(a)` is correct because the graph between `(0-100K)` appears to be a straigh line upto a reasonable approximation. Option `(b)` is correct because area under the curve in the temperature range `(0-100K.)` Option (c) is correct because the graph of `C` versus `T` is constant in the temperature range `(400-500K)` Option `(d)` is correct because in the temperature range `(200-300K)` specific heat capacity increases with temperature. |
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| 7. |
During an experiment, an ideal gas is found to obey an additional law `VP^2=constant,` The gas is initially at a temperature T, and volume V. When it expands to a volume `2V,` the temperature becomes……. |
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Answer» `PV=RT` (ideal gas equation) `implies P=(RT)/(V)` (i) Given that `VP^(2)=`constant (ii) From Eqs. (i) and (ii), `Vxx(R^(2)T^(2))/(V^(2))=`constant `(T^(2))/(V)` =Constant `(T_(1)^(2))/(V_(1))=(T_(2)^(2))/(V_(2))` `implies T_(2)=T_(1)sqrt((V_(2))/(V_(1)))=Tsqrt((2V)/(V))=sqrt(2)T`. |
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| 8. |
Two non-reactive monoatomic ideal gases have their atomic masses in the ratio `2:3.` The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is `4:3.` The ratio of their densities isA. `1:4`B. `1:2`C. `6:9`D. `8:9` |
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Answer» Correct Answer - D `PV=nRT=(m)/(M)RT` `implies PM=pRT` `(rho_(1))/(rho_(2))=(P_(1)M_(1))/(P_(2)M_(2))=(P_(1)/(P_(2)))xx((M_(1))/(M_(12)))=(4)/(3)xx(2)/(3)=(8)/(9)` Here `rho_(1)` and `rho_(2)` are the densities of gases in the vessel containing the mixture. |
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| 9. |
An ideal gas is taken through the cycle `AtoBtoCtoA,` as shown in the figure, If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process CtoA is A. `-5J`B. `-10J`C. `-15J`D. `-20J` |
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Answer» Correct Answer - A For cyclic process, `Q_("cyclic")=W_(AB)+W_(BC)+W_(CA)` `=10J+0+W_(CA)=5J implies W_(CA)=-5J` |
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| 10. |
An ideal monoatomic gas is taken round the cycle ABCDA as shown in the P-V diagram. The work done during the cycle is A. `PV`B. `2PV`C. `1//2`D. zero |
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Answer» Correct Answer - A Work done during the cycle = area enclosed in the curve `=(2P-P)(2V-V)=PV` |
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| 11. |
The curves A and B in the figure shown P-V graphs for an isothermal and an adiabatic process for an idea gas. The isothermal process is represented by the curve A. |
| Answer» The slope of `p-V` curve is more for adiabatic process than for isothermal process. From the graph it is clear that slope for `B` is greater than the slope for `A` . | |
| 12. |
An ideal Black-body at room temperature is thrown into a furnace. It is observed thatA. initially it is the darkest body and later the brightestB. it is the darkest body at all timesC. it cannot be distinguished at all timesD. initially it is the darkest body and later it cannot be distinguished |
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Answer» Correct Answer - A When the temperature of black body becomes equal to the temperature of furnace, it will radiate maximum energy, so it will be brightest. Initially it will absorb all radiations, so it will be darkest. |
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| 13. |
Which of the following graphs correctly represents the variation of `beta=-(dV//dP)/V` with P for an ideal gas at constant temperature?A. B. C. D. |
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Answer» Correct Answer - A `(PV)=`constant Deifferentiating, `(PdV)/(dP)=-V` `beta=-((1)/(V))((dV)/(dP))=((1)/(P)) implies betaxxP=1` Therefore, the graph between `beta` and `P` will be a rectangular hyperbola. `(a)` is the correct option. |
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| 14. |
A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system isA. `4RT`B. `15RT`C. `9RT`D. `11RT` |
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Answer» Correct Answer - D The internal energy of `n` moles of a gas is `u=(1)/(2)nFRT` where `F=`number of degrees of freedom. The internal energy of `2` moles of oxygen at temperature `T` is `u_(1)=(1)/(2)xx2xx5RT=5RT` , (`F=5` for oxygen molecule) Total internal energy `=u_(1)+u_(2)=11RT` |
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| 15. |
From the following statements concerning ideal gas at any given temperature T, select the correct one (s)A. The coefficient of volume expansion at constant pressure is the same for all ideal gases.B. The average translational kinetic energy per molecule of oxygen gas is `3kT`, k being the Boltzmann constant.C. The mean free path of molecules increases with decrease in pressure.D. In a gaseous mixture, the average translational kinetic energy of the molecules of each component is different. |
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Answer» Correct Answer - A::C For `1` mole of an ideal gas `pV=RT` (i) at constant pressure: `PdV=RdT` From Eqs. (i) and (ii), we get (ii) `(dV)/(V)=(dT)/(T)` The coefficient of volume expansion at constant pressure is given by `(dV)/(VdT)=(1)/(T)` same for all gases at same temperature. The average translational kinetic energy per molecule is `(3//2)kT` and not `3kT` . With decrease in pressure, volume of the gas increases so its mean free path increases. [Option `(c)`] The average translational kinetic energy of the molecules is independent of their nature, so each component of the gasesous mixture will have the some value of average translational kinetic energy. |
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| 16. |
During the melting of a slab of ice at 273K at atmospheric pressure,A. positive work is done by the ice-water system on the atmosphereB. positive work is done on the ice-water system by the atmosphereC. the internal energy of the ice-water system increasesD. the internal energy of the ice-water system decreases |
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Answer» Correct Answer - B::C There is a decreases in volume during melting of an ice slab at `273K` . Therefore, negative work is done on the by ice water system on the atmosphere or positive work is done on the ice-water system by the atomsphere. Hence option `(b)` is correct. Second, heat is absorbed during melting (i.e., `dQ` is positive) and as we have seen, work done by icewater system is negative `(dW "is negatvie".)` Therefore, from the first law of thermodynamics, `dU=dQ-dW`, with change in internal energy of ice-water system, `dU` will be positive or internal energy will increase. |
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| 17. |
A spherical body of area A and emissivity `e=0.6` is kept inside a perfectly black body. Total heat radiated by the body at temperature TA. `0.4sigmaAT^4`B. `0.8sigmaAT^4`C. `0.6sigmaAT^4`D. `1.0sigmaAT^4` |
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Answer» Correct Answer - C Heat radiated at temperature `T` : `Q=esigmaAT^(4)=0.6sigmaAT^(4)` |
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| 18. |
At a given temperature, the specific heat of a gas at constant pressure is always greater than its specific heat at constant volume. |
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Answer» `C_(p)gtC_(v)` This is because at constant pressure when heat is supplied to the gas for increasing temperature, some heat is used up in doing work for increasing volume. |
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| 19. |
A fixed thermally conducting cylinder has a radius R and height `L_0`. The cylinder is open at the bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is `P_0`. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then beA. `p_0`B. `(p_0)/(2)`C. `(p_0)/(2)+(Mg)/(piR^2)`D. `p_0/2-(Mg)/(piR^2)` |
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Answer» Correct Answer - A Since it is open from top, pressure will be `p_(0)` Therefore, option `(a)` is correct. |
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| 20. |
The root mean square (r.m.s) speed of oxygen molecules `(O_2)` at a certain temperature T (degree absolute) is V. If the temperature is doubled and oxygen gas dissociates into atomic oxygen, the r.m.s speed remains unchanged. |
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Answer» `v_("rms")=sqrt((3RT)/(M))` `v_("rms")propsqrt((T)/(M))` When oxygen gas dissociates into atomic oxygen, its atomic mass `M` will become half. Temperature is doubled. So, from Eq. (i) `v_("rms")` will become two times. |
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| 21. |
The average translational energy and the rms speed of molecules in a sample of oxygen gas at `300K` are `6.21xx10^(-21)J` and `484m//s`, respectively. The corresponding values at `600K` are nearly (assuming ideal gas behaviour)A. `12.42xx10^(-21)J, 968 m//s`B. `8.78xx10^(-21)J, 684 m//s`C. `6.21xx10^(-21)J, 968m//s`D. `12.42xx10^(-21)J, 684m//s` |
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Answer» Correct Answer - D The formula for average kinetic energy is `bar((KE))=(3)/(2)KT` `:. (bar((KE))_(600K))/(bar((KE))_(300K))=(600)/(300)` `implies bar((KE))_(600K)=2xx6.21xx10^(-21)J=12.42xx10^(-21)J` Also the formula for rms velocity is `C_("rms")=sqrt((3KT)/(m))` `:. ((C_("rms"))_(600K))/((C_("rms"))_(300K))=sqrt((600)/(300))` `implies (C_("rms"))_(600K)=sqrt(2)xx484=684m//s`. |
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| 22. |
A black body of temperature T is inside chamber of `T_0` temperature initially. Sun rays are allowed to fall from a hole in the top of chamber. If the temperature of black body (T) and chamber `(T_0)` remains constant, then A. Black body will absorb radiation.B. Black body will absorb less radiation.C. Black body will emit more energy.D. Black body will emit energy equal to energy absorbed by it. |
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Answer» Correct Answer - A::D Since the sun rays fall on the black body, it will absorb radiations and since its temperature is constant, it will emit radiations. The temperature is will remain same only when energy emitted is equal to energy absorbed. |
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| 23. |
The temperature of an ideal gas is increased from 120K to 480K. If at 120K the root-mean-squre velocity of the gas molecules is v, at 480K it becomesA. `4v`B. `2v`C. `v//2`D. `v//4` |
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Answer» Correct Answer - B `v_("rms")=sqrt((3RT)/(M))` `v_("rms")propsqrt(T)` When temperature is increased from `120K` to `480K` (i.e., four times), the root mean square speed will become `sqrt(4)` or `2` times, `i.e., `2v` . |
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| 24. |
Statement-1: The total translational kinetic energy of fall the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume because. Statement-2: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.A. If both assertion and reason are true and the reason is correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true, but the reason is false.D. If assertion is false, but the reason is true. |
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Answer» Correct Answer - B Total translational kinetic energy `=(3)/(2)nRT=(3)/(2)pV=1.5pV` |
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| 25. |
An ideal gas is initially at temperature T and volume V. Its volume is increased by `DeltaV` due to an increase in temperature `DeltaT,` pressure remaining constant. The quantity `delta=(DeltaV)/(VDeltaT)` varies with temperature asA. B. C. D. |
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Answer» Correct Answer - C We know that `V//T=`constant `(V+DeltaV)/(T+DeltaT)=(V)/(T)` or `VT=TDeltaV=VT+VDeltaT` or `TDeltaV=VDeltaT` or `(DeltaV)/(VDeltaT)=(1)/(T)` This is representeed by graph. |
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| 26. |
An ideal gas with pressure P, volume V and temperature T is expanded isothermally to a volume 2V and a final pressure `P_i,` If the same gas is expanded adiabatically to a volume 2V, the final pressure `P_a.` The ratio of the specific heats of the gas is 1.67. The ratio `(P_a)/(P_1)` is ....... |
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Answer» For isothermal expansion, `PxxV=p_(i)2V` implies `P_(1)=(P)/(2)` For adiabatic expansion, `PV^(y)=P_(a)xx(2V)^(y) implies P_(a)=(P)/(2^(y))=(P)/(2^(1.67))` `:. (P_(a))/(P_(i))=(P)/(2^(1.67))xx(2)/(P)=(1)/(2^(0.67))` |
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| 27. |
A container of volume `1m^3` is divided into two equal parts by a partition. One part has an ideal gas at 300K and the other part is vacuum. The whole system is thermally isolated from the surroundings. When the partition is removed, the gas expands to occupy the whole volume. Its temperature will now be ....... |
| Answer» This is a case of free expansion. Here `DeltaQ=0` and `DeltaW=0` , because gas is doing work on vacuum. So `DeltaU=0` . Hence no change in temperature, it will remain `300K` . | |
| 28. |
An ideal gas is expanding such that `PT^2=constant.` The coefficient of volume expansion of the gas is-A. `1/T`B. `2/T`C. `3/T`D. `4/T` |
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Answer» Correct Answer - C `pT^(2)=`constant `((nRT)/(V))T^(2)=`constant `T^(3)V^(1)=`constant Differentiating the equation, we get `(3T^(3))/(V)dT-(T^(3))/(V(3))dV=0` `3dT=(T)/(V)dV` From the equation, `dV=VgammadT` `gamma=` coefficient of volume expansion of gas `=dV//VdT` . From Eq. (i) , `gamma=(dV)/(VdT)=(3)/(T)` |
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| 29. |
Two moles of ideal helium gas are in a rubber balloon at `30^@C.` The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to `35^@C.` The amount of heat required in raising the temperature is nearly (take R `=8.31 J//mol.K`) |
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Answer» Correct Answer - d `DeltaQ=nC_(p)DeltaT` `=2((f)/(2)R+R)DeltaT` `2=[(3)/(2)R+R]xx5` `=2xx(5)/(2)xx8.31xx5=208J` |
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| 30. |
2kg of ice at `20^@C` is mixed with 5kg of water at `20^@C` in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water & ice are `1kcal//kg//^@C` and `0.5` `kcal//kg//^@C` while the latent heat of fusion of ice is `80kcal//kg`A. `7kg`B. `6kg`C. `4kg`D. `2kg` |
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Answer» Correct Answer - B Heat required to convert `5kg` of water at `20^(@)C` to `5kg` of water at `0^(@)C=mC_(omega)DeltaT=5xx1xx20=100Kcal` . How much ice at `0^(@)C` will convert into water at `0^(@)C` for giving another `80Kcal` of heat `Q=mL` `implies 80=mxx80 implies m=1kg` Therefore, the amount of water at `0^(@)C=5kg+1kg=6kg` . Thus, at equilibrium we have (`6kg` water at `0^(@)C+1kg` ice at `0^(@)C`) |
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| 31. |
One mole of a mono-atomic ideal gas is mixed with one mole of a diatomic ideal gas. The molar specific heat of the mixture at constant volume is …… |
| Answer» `C_(v)=(n_(1)C_(v_(1))+n_(2)C_(v_(2)))/(n_(1)+N_(2))=((1)((3)/(2)R)+(1)((5)/(2)R))/(1+1)=2R` | |
| 32. |
When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas isA. `2/5`B. `3/5`C. `3/7`D. `5/7` |
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Answer» Correct Answer - D The desired fraction is `f=(DeltaU)/(DeltaQ)=(nC_(v)DeltaT)/(nC_(p)DeltaT)=(C_(v))/(C_(p))=(1)/(gamma)` `f=(5)/(7)` , `("as" gamma=(7)/(5))` |
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| 33. |
At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be `1930 m//s`. The gas isA. `H_2`B. `F_2`C. `O_2`D. `CI_2` |
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Answer» Correct Answer - A `v_("rms")=sqrt((3RT)/(M))` Room temperature, `T~~300K` `1930=sqrt((3xx8.31xx10^(3)xx300)/(M))` `M=2g` or the gas is `H_(2)` . |
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| 34. |
If one mole of a monatomic gas `(gamma=5/3)` is mixed with one mole of a diatomic gas `(gamma=7/5),` the value of gamma for mixture isA. `1.40`B. `1.50`C. `1.53`D. `3.07` |
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Answer» Correct Answer - B `gamma_(1)=5//3` means gas is monatomic or `C_(v_(1))=(3)/(2)R` `gamma_(2)=7//5` means gas is diatomic or `C_(v_(2))=5//2R` `C_(v)` (of the mixture) `=(n_(1)C_(v_(1))+n_(2)C_(v_(2)))/(n_(1)+n_(2))=((1)((3)/(2)R)+(1)((5)/(2))R)/(1+1)=2R` `C_(p)` (of the mixture) `=C_(v)+R=3R` `gamma_("mixture")=(C_(p))/(C_(v))=(3R)/(2R)=1.5` |
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