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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.
| 151. |
If `C_(1), C_(2), C_(3) …..` represent the speeds on `n_(1), n_(2) , n_(3)…..` molecules, then the root mean square speed isA. `((n_(1)C_(1)^(2)+n_(2)C_(2)^(2)+n_(3)C_(3)^(2)+...........)/(n_(1)+n_(2)+n_(3)+.......))^(1//2)`B. `((n_(1)C_(1)^(2)+n_(2)C_(2)^(2)+n_(3)C_(3)^(2)+...........)^(1//2))/(n_(1)+n_(2)+n_(3)+.......)`C. `((n_(1)C_(1)^(2))^(1//2))/(n_(1))+((n_(2)C_(2)^(2))^(1//2))/(n_(2))+((n_(3)C_(3)^(2))^(1//2))/(n_(3))+......`D. `[((n_(1)C_(1)+n_(2)C_(2)+n_(3)C_(3)+....)^(2))/((n_(1)+n_(2)+n_(3)+....))]^(1//2)` |
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Answer» Correct Answer - A Root mean square speed `[ (n_(1)c_(1)^(2)+n_(2)c_(2)^(2)+n_(3)c_(3)^(2))/(n_(1)+n_(2)n_(3)+....)]^(1//2)` |
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| 152. |
At a definit temperature (T), the distribution of speeds is given by the curve. In the curve points A, B and C indicates the speeds corresponding to : A. most probable, average and root mean square speedsB. average, root mean square and most probable speedsC. root mean square, average and most probable speedsD. most probable, root mean square and average speeds |
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Answer» Correct Answer - a (a) `U_("rms")=sqrt((2RT)/(M)) " "U_("avg")=sqrt((8)/(pi)(RT)/(M)) " "U_("mp")=sqrt((2RT)/(M))` `underset (1.224)(U_("rms"))" : "underset(1.128)(U_("avg"))" : "underset(1)(U_("mp"))` `U_("rms")gtU_("avg")gtU_("mp")` |
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| 153. |
The Ne atom has 10 times the mass of `H_(2)` molecule. Which of the following statements is true? I. At `25^(@)C` both of them have the same kinetic energy. II. Ten moles of `H_(2)` would have the same volume as 1 mole of Ne at same temp. and pressure. III. One mole of Ne exerts the same pressure as one mole of `H_(2)` at STP. IV. A `H_(2) ` molecule travels 10 times faster than Ne atom at same temperature. V. At STP, one litre of Ne has 10 times the density of 1 litre of `H_(2).`A. II, IV, VB. I, III, VC. I, II, IIID. I, II |
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Answer» Correct Answer - b (b) K.E. is temprature dependent same moles would have same volume at same P and T `PV=nRT` `d=(PM)/(RT)` |
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| 154. |
Which of the following statements about kinetic energy (K.E.) is true?A. All objects moving with the same velocity have the same K.E.B. The K.E. of a body will quadruple if its velocity doublesC. As the velocity of a body increases, its K.E. decreasesD. The K.E. of a body is independent of its mass |
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Answer» Correct Answer - b (b) `KE=(1)/(2)Mv^(2)` |
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| 155. |
n mol of `N_(2)` and 0.6 mol of Ar are enclosed in a vessel of capacity 2L at 1 atm and `27^(@)C`?A. 0.3B. 0.1C. 0.03D. 0.06 |
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Answer» Correct Answer - c |
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| 156. |
The rms velocity of hydrogen is `sqrt(7)` times the rms velocity of nitrogen. If `T` is the temperature of the gas, thenA. `T(H_(2))=T(N_(2))`B. `T(H_(2)) gt T(N_(2))`C. `T(H_(2)) lt T(N_(2))`D. `T(H_(2))= sqrt(7) T(N_(2))` |
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Answer» Correct Answer - C `sqrt((3RT_(H_(2)))/(2))=sqrt(7) sqrt((3RT_(N_(2)))/(28)), :. T_(N_(2))=2T_(H_(2))` |
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| 157. |
The rms velocity of hydrogen is `sqrt(7)` times the rms velocity of nitrogen. If T is the temperature of the gas, then:A. `T_((H_(2)))=T_((N_(2)))`B. `T_((H_(2)))gtT_((N_(2)))`C. `T_((H_(2)))ltT_((N_(2)))`D. `T_((H_(2)))=sqrt(7)T_((N_(2)))` |
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Answer» Correct Answer - c |
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| 158. |
Which of the following statements is not true about the effect of a n increase in temperature on the distribution of molecular speeds of an ideal gas?A. The area under the curve remains same even at the higher temperature.B. The distribution pattern becomes more uniform.C. Fraction of molecules with speed greater than a particular high speed will increases.D. The fraction of molecules having most probable |
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Answer» Correct Answer - d |
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| 159. |
The root mean square speed of hydrogen is `sqrt(5)` times than that of nitrogen. If T is the temperature of the gas, then :A. `T_(H_(2))=T_(N_(2))`B. `T_(H_(2)) gt T_(N_(2))`C. `T_(H_(2))lt T_(N_(2))`D. `T_(H_(2)) lt sqrt(7)T_(N_(2))` |
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Answer» Correct Answer - 3 `V_(rms)=sqrt((3RT)/(M))=((V_(rms))_(H_(2)))/((V_(rms))N_(2))=sqrt((T_(H_(2)))/(M_(H_(2)))xx(M_(N_(2)))/(T_(N_(2)))),` `(V_(rms))H_(2)=sqrt(5)(V_(rms))N_(2)` `((V_(rms))H_(2))/((V_(rms)N_(2)))xxsqrt(5)=sqrt((T_(H_(2)))/(M_(H_(2)))xx(M_(N_(2)))/(T_(N_(2)))),` `(sqrt(5))/(1)=sqrt((T_(H_(2)))/(T_(N_(2)))xx14)` `5=(T_(H_(2)))/(T_(N_(2)))xx14` `T_(N_(2))xx5=T_(H_(2))xx14` `T_(N_(2))gtT_(H_(2))` |
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| 160. |
The root mean square speed of hydrogen is `sqrt(5)` times than that of nitrogen. If T is the temperature of the gas, then :A. `T_(H_(2))=T_(N_(2)`B. `T_(H_(2))gtT_(N_(2)`C. `T_(H_(2))ltT_(N_(2)`D. `T_(H_(2))=sqrt(7)T_(N_(2)` |
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Answer» Correct Answer - c (c) `U_("rms")=sqrt((3RT)/(Mw))=((U_("rms"))_(H_(2)))/((U_("rms"))_(N_(2)))=sqrt(T_(H_(2))/(M_(H_(2)))xxM_(N_(2))/(T_(N_(2)))),` `(U_("rms"))_(H_(2))=sqrt(5)(U_("rms"))_(N_(2))` `:.((U_("rms"))_(H_(2)))/((U_("rms"))_(H_(2)))xxsqrt(5)=sqrt(T_(H_(2))/(T_(N_(2)))xx(28)/(2))` `=(sqrt(5))/(1)=sqrt(T_(H_(2))/(T_(N_(2)))xx14)` `=5=T_(H_(2))/(T_(N_(2)))xx14` `T_(N_(2))xx5=T_(H_(2))xx14` `:. " "T_(N_(2))gtT_(H_(2))` |
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| 161. |
Which one of the following statement is not true about the effect of an increase in temperature on the distribution of molecular speed of gas ? .A. The most probable speed increasesB. The fraction of the molecules with the most probable speed increasesC. The distribution becomes broaderD. The area under the distribution curve remaiins the same as under the lower temperature |
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Answer» Correct Answer - B Most probable velocity increase and fraction of molecule possessing most probable velocity decrease. |
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| 162. |
Root mean square velocity of a particle is v at pressure P. If pressure is increased two times, then the r.m.s. velocity becomesA. 1200KB. 900KC. 600KD. 150 K |
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Answer» Correct Answer - A `V_(rms)=sqrt((3RT)/(M))implies V_(rms)=sqrt(T)` Given `V_(1)=V,T_(1)=300K, V_(2)=2V,T_(2)=?` `(V_(1))/(V_(2))=sqrt((T_(1))/(T_(2))),((V)/(2V))^(2)=(300)/(T_(2))impliesT_(2)=300xx4=1200K` |
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| 163. |
At `27^(@)C`, the ratio of rms velocities of ozone to oxygen isA. `sqrt(3//5)`B. `sqrt(4//3)`C. `sqrt(2//3)`D. `0.25` |
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Answer» Correct Answer - C `(u_(O_(3)))/(u_(O_(2)))=sqrt((M_(O_(2)))/(M_(O_(3))))=sqrt((32)/(48))=sqrt((2)/(3))` |
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| 164. |
If the `v_(rms)` is `30R^(1//2)` at `27^(@)C` then calculate the molar mass of gas in kilogram.A. 1B. 2C. 4D. `0.001` |
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Answer» Correct Answer - D `v_(rms)=sqrt((3RT)/(M))` `sqrt(30^(2)R)=sqrt((3RT)/(M))` `implies 30 xx 30R =(3Rxx300)/(M)` `implies M = (3 xx 300)/(30xx30)=1 gm = 0.001kg`. |
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| 165. |
A mixture of hydrogen and oxygen gas weighing `0.76g` has volume 2 liter temperature `300K` and pressure `100 kPa` Calculate masses of hydrogen and oxygen in gas mixture Assume ideal nature . |
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Answer» `V =2litre = 2 xx 10^(-3) m^(3),` `P =100kPa = 100 xx 10^(3) Pa` `w_(H_2) + w_(O_2) = 0.76g` Also `n = [n_(H_2) + n_(O_2)] = (PV)/(RT)` ` = (100xx10^(3) xx 2xx 10^(-3))/(8.314 xx 300) = 0.08` or `(w_(H_2))/(2) + (w_(O_2))/(32) = 0.08` or `16w_(H_2) + w_(O_2) = 2.56` By eqns (i) and (ii) `w_(H_2) = 0.12 g, w_(O_2) = 0.64 g` . |
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| 166. |
A definite quatity of an ideal gas has a volume of 30 liter at `20^(@)C` The gas is first compressed at `20^(@)C` until the pressure has doubled and then the temperature is raised to `100^(@)C` keeping pressure constant Determine the final volume of gas . |
| Answer» Correct Answer - `19.1 litre` | |
| 167. |
Which of the following statements are correct?A. It is not possible to compress a gas at temperature below `T_(c)`B. At a temperature below `T_(c)`, the molecules are close enough for the attractive forces to act and condensation occursC. No condensation takes place above `T_(c)`D. Due to higher kinetic energy of gas molecules above `T_(c)`, it is considered as super critical fluid |
| Answer» Correct Answer - B::C::D | |
| 168. |
The given represents the variation of Z (compressibility factor `=(V)/(nRT)` versus P, for three real gas A,B and C . Identify the only incorrect statement. A. For the gas A, `a=0` and its dependence on P is linear at all pressureB. For the gas B, `b=0` and its dependence on P is linear at all pressureC. For the gas C, which is typical real gas for which neither a nor `b=0`. By knowing the minima and the point of intersection , with `z=1`, a and b can be calculatedD. At high pressure , the slope is positive for all real gases |
| Answer» Correct Answer - B | |
| 169. |
A monotomic ideal gas undergoes a process in which the ratio of p to V at any instant is constant and equals to 1. what is the molar heat capacity of the gas?A. `(4R)/(2)`B. `(3R)/(2)`C. `(5R)/(2)`D. 0 |
| Answer» Correct Answer - C | |
| 170. |
A monoatomic ideal gas undergoes a process in which the ration of `P` to `V` at any instnat is constnat and equal to unity The molar heat capacity of gas is .A. `2.0 R`B. `1.5 R`C. `2.5 R`D. `0` |
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Answer» Let `P,V` be the pressure and volume of gas at temperature `T` `P_(1)V_(1) =RT` `P_(2) (V_(1)) +dV =R (T +1)` `:. P_(2)^(2) =RT +R ( :. (P_(2))/(V_(1)+dV) =1)` `:. 22((deltaP_(2))/(delta))_(V) =R` `((deltaP_(2))/(deltaT))=(R)/(2)` `:. C =C_(v) + (deltaP)/(deltaT) = (3R)/(2) + (R)/(2) =2R` . |
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| 171. |
4.4 g of a gas at STP occupies a volume of 2.24 L. The gas can be :A. `O_(2)` and `CO_(2)`B. `CO`C. `NO_(2)`D. `CO_(2)` |
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Answer» Correct Answer - D Molecular weight `=(mRT)/(Pv)=(4.4 xx.082xx273)/(1 xx 2.24 )= 44` So the gas should be `CO_(2)` |
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| 172. |
If `P` is the pressure of gas, then the kinetic energy per unit volume of the gas is .A. `P//2`B. `P`C. `3P//2`D. `2P` |
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Answer» `K.E - (3RT)/(2) = (3PV)/(2)` `:. (K.E)/(V) = (3P)/(2)` . |
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| 173. |
Vapour ensity of a gas showing endothermic dissociation equilibria on heating .A. Decreases with increase in temperatureB. Increases with increase in temperatureC. Remains same with increase in temperatureD. May increase of derease with increase in temperature |
| Answer» The mo1 wt of gaseous mixture ar equilibrium will be less . | |
| 174. |
A vessel of volume 5 litre contains 1.4 g of nitrogen at a temperature 1800K. The pressure of the gas is 30% of its molecules are dissociated into atoms at this temperature is:A. 4.05 atmB. 2.025atmC. 3.84atmD. 1.92atm |
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Answer» Correct Answer - d |
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| 175. |
A barometer is an instrument that is used for the measurement of pressure The construction of the barometer is as follows: A thin narrow calibrated capillary tube is filled to the brim with a liquid such as mercury and is inverted into a through filled with the same fluid Now depending on the external atmospheric pressure the level of the inside capillary comes to rest, then the net forces on the column should be balanced Applying force balance we get `P_(atm)xxA=mxxg` (A is the cross-sectional area of the capillary tube) if `rho` is the density of the fluid, then `m=rhoxxgxxh` (h is the height to which mercury has risen in the capillary) Hence `P_(atm)xxA=(rhoxxgxxh)xxA or P_(atm)=rhogh` Faulty Barometer: An ideal barometer will show a correct reading only if the space above the mercury column is vacuum but in case if some gas column is trapped in the space above the mercury column then the barometer is classified as a faulty barometer. The reading of such a barometer will be less than the true pressure For such a faulty barometer `P_(0)A=mg+P_(gas)A` `P_(0)=rhohg+P_(gas)` or `rhogh=P_(0)-P_(gas)` A gas column is trapped between closed end of a tube and a mercury column of length (h) when this tube is placed with its open end upwards the length (h)when this tube is placed with its open end upwards the length of gas column is `(l_(1))` the length of gas column becomes `(l_(2))` when open end of tube is held downwards (as shown in figure) find atmospheric in terms of height of Hg column. (Assume temeperature remains constant) A. `(h(l_(1)+l_(2)))/(l_(2)-l_(1)`B. `(h(l_(2)+l_(1)))/(l_(1)-l_(2)`C. `(l_(1)+l_(2))/(h(l_(2)-l_(1))`D. `(h_(1)l_(1)+h_(2)l_(2))` |
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Answer» Correct Answer - a |
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| 176. |
Calculate van der Waals constansts a and b if critical temperature and critical pressure are `30^(@)C` and 72 atm respectively . |
| Answer» Correct Answer - `a = 0.36 atm litre^(2) mole^(-2), b = 4.28 xx 10^(-3) litre//mole` | |
| 177. |
समीकरण `C_(P) -C_(V) = R,` में R का अर्थ होता है :A. Work done per mole per KelvinB. Heat absorbed per mole per KelvinC. Heat released per mole per KelvinD. Work done per mole per degree celcius |
| Answer» Correct Answer - A | |
| 178. |
Pressure (P ) vs. density (D ) curve for an ideal gas at two different temperature `T_(1)` and `T_(2)` is shown below Identify the correct statement about `T_(1)` and `T_(2)`A. `T_(1) gt T_(2)`B. `T _(1) lt T_(2)`C. `T_(1) =T_(2)`D. Cannot be said |
| Answer» Correct Answer - A | |
| 179. |
`I`, `II`, and `III` are three istherms, respectively, at `T_(1)`, `T_(2)`, and `T_(3)`. Temperature will be in order A. `T_(1) =T_(2) =T_(3)`B. `T_(1) lt T_(2) gt T_(3)`C. `T_(1) gt T_(2) gt T_(3)`D. `T_(1) gtT_(2) =T_(3)` |
| Answer» As the temperature increases at constant `P` volume of gas increases . | |
| 180. |
Four particles have speed `2, 3, 4` and 5 cm/s respectively Their `RMS` speed is .A. 3.5 cm/sB. (27/2) cm/sC. `sqrt(54)` cm/sD. `sqrt(54) //2` cm/s |
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Answer» Correct Answer - d (d) `U_("rms")=sqrt((U_(1)^(2)+U_(2)^(2)+U_(3)^(2)+U_(4)^(2))/(4))=sqrt((54)/(2))=(sqrt(54))/(2)` |
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| 181. |
At what temperature, the average speed of gas molecules be double of that at temperature, `27^(@)C`?A. `120^(@)C`B. `108^(@)C`C. `927^(@)C`D. `300^(@)C` |
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Answer» Correct Answer - c |
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| 182. |
At the same temperature and pressure, which of the following will have same kinetic energy per mole as `N_(2)O`?A. HeB. `H_(2)S`C. `CO_(2)`D. `NO_(3)` |
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Answer» Correct Answer - a,b,c,d |
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| 183. |
A gaseous mixture contains 4 molecules with a velocity of 6 cm `sec^(-1)`,5 molecules with a velocity of 2 cm `sec^(-1)` and 10 molecules with a velocity of 3 cm `sec^(-1)`. What is the rms speed of the gas:A. 2.5 cm `sec^(-1)`B. 1.9 cm `sec^(-1)`C. 3.6 cm `sec^(-1)`D. 4.6 cm `sec^(-1)` |
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Answer» Correct Answer - c (c) `U_("ms")=sqrt((4xx(6)^(2)+5(2)^(2)+10(3)^(2))/(19)` `=3.6cm//sec.` |
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| 184. |
The ratio between the root mean square speed of `H_(2)` at `50 K` and that of `O_(2)` at `800 K` isA. 4B. 2C. 1D. `1//4` |
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Answer» Correct Answer - c (c) `U_("ms")=sqrt((3RT)/(Mw))` `(U_(H_(2)))/(U_(O_(2)))=sqrt((50)/(2)xx(32)/(800))=1` |
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| 185. |
Which of the following expression correctly represents the relationship between the average kinetic energy of CO and `N_(2)` molecules at the same temperature?A. `vecE(CO)gtvecE(N_(3))`B. `vecE(CO)ltvecE(N_(2))`C. `vecE(CO)=vec(E)N_(2)`D. Cannot be predicted unless volumes of the gases are given |
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Answer» Correct Answer - c |
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| 186. |
The quantity `(PV)/(k_(B)T)` represents the (` k_(B):`Boltzmann constant)A. number of particles of the gasB. mass of the gasC. number of moles of the gasD. translation energy of the gas |
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Answer» Correct Answer - a (a) `PV=nRT " "R=KxxN_(A)` `PV=nxxN_(A)KT` `nxxN_(A)=(PV)/(KT)=` no. of particles |
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| 187. |
Following graph represents a pressure `(P)` volume `(V)` relationship at a fixed temperature `(T)` for `n` moles of a real gas. The graph has two regions marked `(I)` and `(II)`. Which of the following options is true. A. `Z gt1` in the region `(II)`.B. `Z=1` in the region `(II)`C. `Z=1` for the curveD. `Z` approaches 1 as we move from region `(II)` to region `(I)` |
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Answer» Correct Answer - 4 On moving from region `(II)` to region `(I)`, pressure tends to zero. So, `Z rarr 1`. |
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| 188. |
Oxygen gas generated by the decomposition of potassium chlorate is collected. The volume of oxygen collected at `24^(@)C` and atmospheric pressure of `760 m m Hg` is `128 mL`. Calculate the mass `(` in grams `)` of oxygen gas obtained. The pressure of water vapour at `24^(@)C` is `22.4 m m Hg`.A. `1.36`B. `1.52g`C. `0.163g`D. `1.63g` |
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Answer» Correct Answer - C From the total pressure and the vapour pressure of water we can calculate the partial pressure of `O_(2)`. `P_(O_(2))=P_(tau)-P_(H_(2)O)` `=660-22.4` `=737.6mm Hg` From the ideal gas equation we write. `m=(PVM)/(RT) =((0.974atm)(0.128L)(32.0g//mol))/((0.0821Latm//Kmol)(273+24)K)=0.163g.` |
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| 189. |
What is the total pressure in a 2.00 L container that holde 1.00 g He, g CO and 10.0 g of NO at `27^(@)C`?A. 21.6atmB. 13.2atmC. 1.24atmD. 0.310atm |
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Answer» Correct Answer - b |
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| 190. |
Longest mean free path stands forA. `H_(2)`B. `N_(2)`C. `O_(2)`D. `Cl_(2)` |
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Answer» Correct Answer - A The mean free path , `lambda=(1)/(sqrt(2)pi sigma^(2)N) `or `lambda prop (1)/(a^(2))`, where ` a=` molecular diameter `:. `Smaller the molecular diameter, longest the mean free path Hence, `H_(2)` is the answer. |
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| 191. |
In a rigid contaienr `NH_(3)` is kept at a certain temperature where its presure is found to be `P_(1)`, if dissociated into `N_(2)` and `H_(2)` at these conditions pressure of mixture is found to be `P_(s)`. Find ratio of `P_(2)` of `P_(1)`.A. 4B. 2C. `(1)/(2)`D. `(1)/(4)` |
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Answer» Correct Answer - a |
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| 192. |
When `CO_(2)` under high pressure is released from a fire extinguisher. Particles of soloid `CO_(2)` are formed, despite the low sublimation temperature `(-77^(@)C)` of `CO_(2)` at 1.0atm it is:A. the gas does work pushingback the atmosphere using KE of molecuels and thus lowering the temperatureB. volume of the gas is decreased rapidly hence, temperature is loweredC. both a and bD. none of the above |
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Answer» Correct Answer - a |
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| 193. |
Which of the following change is observed occurs when a substance X is converted from liquid to vapour phase at the standard boiling point? I. Potential energy of the system cecreases II. The distance between molecules increases III.The average kinetic energy of the molecules in both phases are equalA. I onlyB. II onlyC. III onlyD. II and III only |
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Answer» Correct Answer - d (d) Due to phase transfer distance between particle changes, because temperature remains constant, so KE remains constant. |
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| 194. |
A helium atom is two times heavier than a hydrogen molecule. At `298 K`, the average kinetic energy of a helium atom isA. two times that of hydrogen moleculesB. same as that jof hydrogen moleculesC. four times that of hydrogen moleculesD. half that of hydrogen molecules |
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Answer» Correct Answer - b |
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| 195. |
Helium atom is two times heavier than a hydrogen molecule at 298 K, the average kinetic energy of helium isA. two times that of hydrogen moleculesB. same as that jof hydrogen moleculesC. four times that of hydrogen moleculesD. half that of hydrogen molecules |
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Answer» Correct Answer - b |
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| 196. |
A helium atom is two times heavier than a hydrogen molecule. At `298 K`, the average kinetic energy of a helium atom isA. Two times that of a hydrogen moleculeB. Same as that of a hydrogen moleculeC. Four times that of a hydrogen moleculeD. Half that of a hydrogen molecule |
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Answer» Correct Answer - B `E prop T` `(E_(He))/(H_(2))=(T_(He))/(T_(H_(2)))` , He and `H_(2)` at same temperature |
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| 197. |
The temperature at which a real gas obeys the ideal gas laws over a wide range of pressure is calledA. Critical temperatureB. Inversion temperatureC. Boyle temperatureD. Reduced temperature |
| Answer» Correct Answer - C | |
| 198. |
If a gas expands at constant temperature , it indicates thatA. Kinetic energy of molecules remains the sameB. Number of the molecules of gas increasesC. Kinetic energy of molecules decreasesD. Pressure of the gas increase |
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Answer» Correct Answer - A Kinetic energy of gaseous molecules depend on temperature only. |
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| 199. |
Statement-1: Gas with lower molar mass will effuse or diffuse faster. Statement-2: Total kinetic Energy of any gas depednds upon its molar mass.A. Statement-I is True, Statement-II is True : Statement-II is a correct explanation for Statement-IB. Statement-I is True, Statement-II is True : Statement-II is `NOT` a correct explanation for Statement-IC. Statement-I is True, Statement-II is False.D. Statement-I is False, Statement-II is True. |
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Answer» Correct Answer - b |
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| 200. |
Pressure of 1 g ideal gas X at 300 K is 2 atm. When 2 g of another gas Y is introduced in the same vessel at same temperature, the pressure become 3 atm. The correct relationship between molar mass of X and Y is :A. `M_(Y)=2 M_(X)`B. `M_(Y)=4 M_(X)`C. `M_(X)=4 M_(Y)`D. None of these |
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Answer» Correct Answer - b (b) `PxxN` `(P_(x))/(P_(y))=(n_(x))/(n_(y))` ` (2)/(1)=(1//m_(x))/(2//m_(y))` `m_(y)=4m_(x)` |
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