InterviewSolution
Saved Bookmarks
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.
| 651. |
An ideal gas is heated at contant volume until its pressure doubles. Which one of the following statements is correct?A. The mean speed of the molecule doubles.B. The number of molecules doubles.C. The mean square speed of the molecules doubles.D. The number of molecules per unit volume doubles. |
|
Answer» Correct Answer - C Assume the mass of the gas is unchanged. Use the kinetic theory equation to relate the various factors to the pressure chage at constant volume. |
|
| 652. |
If the volume of a gas is doubled at constant pressure, the average translational kinetic energy of its molecules willA. be doubledB. remain the sameC. increase by a factorD. become fore times |
|
Answer» Correct Answer - A If volume is doubled at constant pressure, then absolute temperature of the gas is doubled. So, average translational kinetic energy is also doubled. |
|
| 653. |
A piece of ice at `0^@C` is dropped into water at `0^@C`. Then ice willA. meltB. be converted into waterC. not meltD. partially melt |
| Answer» Correct Answer - C | |
| 654. |
Rate of cooling of body is `0.5^(@)C`/min, when the system is `50^(@)C` above the surroundings. When a system is `30^(@)C` above the surroundings, the rate of cooling will beA. `0.3^(@)C`/minB. `0.6^(@)C`/minC. `0.7^(@)C`/minD. `0.4^(@)C`/min |
|
Answer» Correct Answer - A `(R_(2))/(R_(1))=(K(theta_(2)-theta_(1)))/(K(theta_(1)-theta_(0)))=(30)/(50)` `R_(2)=(3)/(5)R_(1)` `=(3)/(5)xx0.5=0.3.^(@)C//`min. |
|
| 655. |
If gas molecules undergo, inelastic collision with the walls of the containerA. the temperature of the gas will decreaseB. the pressure of the gas will increseC. neither the temperature nor the pressure will changeD. the temperature of the gas will increase |
| Answer» Correct Answer - C | |
| 656. |
Which of the following at `100^@C` produces most sever burns ?A. Hot airB. WaterC. SteamD. Oil |
| Answer» Correct Answer - C | |
| 657. |
What enrgy transformation takes place when ice is converted into waterA. Heat energy to kinetic energyB. kinetic energy to heat energyC. Heat energy to latent heatD. Heat energy to potential energy |
| Answer» Correct Answer - D | |
| 658. |
A block of ice falls from certain height and completely melts. If only 3/4th of the enrgy is absorbed by the block. The height of the fall should be (L=363SI units and `g=10ms^(-2)`)A. 48.4mB. 84.4mC. 88.4mD. 44.8m |
|
Answer» Correct Answer - A `(3)/(4)mgh=JmL` |
|
| 659. |
Find the work done by the gas in the process `ABC`. A. `(3)/(2) P_(0) V_(0)`B. `(5)/(2) P_(0) V_(0)`C. `(7)/(2) P_(0) V_(0)`D. `4 P_(0) V_(0)` |
|
Answer» Correct Answer - C `W_(AB) = -(P_(0)V_(0)+(P_(0)V_(0))/(2)) = -3/2 P_(0)V_(0)` `W_(AB) = (2P_(0))(2V_(0))+(P_(0)(2V_(0)))/(2) = 5P_(0)V_(0)` `W_(AB) = 7/2 P_(0)V_(0)`. |
|
| 660. |
In a mechanical refrigerator the low temperature coils are at a temperature of `-23^(@)C` and the compressed gas in the condenser has a temperature of `27^(@)C`. The theoretical coefficient of performance isA. 5B. 8C. 6D. 6.5 |
|
Answer» Correct Answer - A `beta=(Q_(2))/(W)=(T_(2))/(T_(1)-T_(2))` |
|
| 661. |
In a mechanical refrigerator the low temperature coils are at a temperature of `-23^(@)C` and the compressed gas in the condenser has a temperature of `27^(@)C`. The theoretical coefficient of performance isA. `5`B. `8`C. `6`D. `6.5` |
|
Answer» Correct Answer - A Coefficient of performance `K = (T_2)/(T_1-T_2) ` `((273-23))/((273+27)-(273-23)) = (250)/(300-250) = (250)(20)`. |
|
| 662. |
The difference between the principal specific heats of nitrogen is 300 J/kg K and ratio of the two specific heats is 1.4. then `C_P` isA. 1050 J/kg KB. 750 J/kg KC. 650 J/kg KD. 150 J/kg K |
|
Answer» Correct Answer - A `C_(p)=C_(V)+300=750+300=1050K`. |
|
| 663. |
If the difference between the principal specific heats of nitrogen is 300 J/kg K and ratio of specific heat is 1.4 then `c_(v)` will beA. 1050 J/kg KB. 250 J/kg KC. 750 J/kg KD. 150 J/kg K |
|
Answer» Correct Answer - C `C_(p)-C_(V)=300` `(C_(p))/(C_(V))-1=(300)/(C_(V))` `1.4-1=(300)/(C_(V))` `C_(V)=(300)/(0.4)=(3000)/(4)=750` |
|
| 664. |
The difference between the specific heat is 600 J/kg K. the ratio of their specific heats is 1.6. the value of `c_(p)` is . . ..J/kg K.A. 600B. 1600C. 1000D. 40 |
|
Answer» Correct Answer - B `c_(p)-c_(V)=600c_(p)` `=1.6c_(V)` `1.6c_(V)-c_(V)=600,0.6c_(V)=600` `c_(V)=1000J//kg` K `therefore c_(p)=1.6c_(V)` `=1600J//kg` K |
|
| 665. |
The ratio of specific heats of a gas is 1.4. if the value of `C_(V)` is 20.8 J/mol K, then the `C_(P)` isA. 3.93 cal/mol KB. 4.93 cal/mol KC. 5.93 cal/mol KD. 6.93 cal/mol K |
|
Answer» Correct Answer - D `(C_(p))/(C_(V))=1.4` `C_(p)=1.4C_(V)=1.4xx20.8` `=29.12J//mol` K `C_(p)=(29.12)/(4.2)=6.93` cal/mol K |
|
| 666. |
Calculate the RMS velocity of molecules of a gas of which the ratio of two specific heats is 1.42 and velocity of sound in the gas is 500 m/s.A. 727 m/sB. 527 m/sC. 927 m/sD. 750 m/s |
|
Answer» Correct Answer - A `V_(S)=sqrt((gamma)/(3))C_(rms)` `=sqrt((1.42)/(3))xxC_(rms)` `500=sqrt(0.47)C_(rms)` `C_(rms)=(500)/(sqrt(0.47))=727m//s`. |
|
| 667. |
The temperature at which rms velocities of nitrogen gas molecules will be doulbed that at `0^(@)C` isA. 273 KB. 546 KC. 136 KD. 1092 K |
|
Answer» Correct Answer - D `C_(1)propsqrt(T_(1)) and C_(2) prop sqrt(T_(2))(C_(2))/(C_(1))=sqrt((T_(2))/(T_(1)))` `therefore T_(2)=(C_(2)^(2))/(C_(1)^(2))T_(1)` `=4xx273=1092K` |
|
| 668. |
The temperature at which mean kinetic energy of an ideal gas molecule will be doubled that at `27^(@)C` isA. `54^(@)C`B. `13.5^(@)C`C. `600^(@)K`D. `150^(@)K` |
|
Answer» Correct Answer - C `K.E_(2) prop T_(1) and KE_(2) prop T_(2)` `(KE_(2))/(KE_(1))=(T_(2))/(T_(1))` `therefore (2KE_(1))/(KE_(1))xxT_(1)=T_(2)` `2xx300=T_(2)` `therefore T_(2)=600K` |
|
| 669. |
If velocities of 3 molecules are 5 m/s, -6m/s and 7 m/s respectively, then their mean squar evelocity in `m^(2)//s^(2)`A. `11m^(2)//s^(2)`B. `36.7m^(2)//s^(2)`C. `6m^(2)//s^(2)`D. `2m^(2)//s^(2)` |
|
Answer» Correct Answer - B `C^(2)=(C_(1)^(2)+C_(2)^(2)+C_(3)^(2))/(3)=(25+36+49)/(3)` `=(110)/(3)=36.66` |
|
| 670. |
If the mean kinetic energy of the molecules of a gas is `(1/3)^(rd)` of its value at `27^(@)C`, then the temperature of the gas will beA. `100^(@)C`B. `-173^(@)C`C. `900^(@)C`D. `627^(@)C` |
|
Answer» Correct Answer - B `KE_(2)=(1)/(3)KE_(1)` `(KE_(2))/(KE_(1))=(T_(2))/(T_(1))" "therefore(1)/(3)=(T_(2))/(T_(1))` `therefore T_(2)=(T_(1))/(3)=(300)/(3)=100K` `T_(2)=100-273=-173^(@)C` |
|
| 671. |
The r.m.s. velocity of the molecules in a gas at `27^(@)C` is 300 m/s. then the r.m.s. velocities o the molecules in the same gas at `927^(@)C` isA. 1200 m/sB. 600 m/sC. 150 m/sD. 75 m/s |
|
Answer» Correct Answer - B `(C_(2))/(C_(1))=sqrt((T_(2))/(T_(1)))=sqrt((1200)/(300))=sqrt(4)` `(C_(2))/(C_(!))=2` `C_(2)=2C_(1)=2xx300=600m//s` |
|
| 672. |
At what temperature is the `K.E`. Of a gas molecules half that of its value at `27^(@)C`A. `13.5^(@)C`B. `150^(@)C`C. `150 K`D. `-123 K` |
|
Answer» Correct Answer - C `(T)/(300) = 1/2 or T = 150K`. |
|
| 673. |
The core of the sun is a plasma. It is at a temperature of the order of `10^(7)` K and contains equal number of protons and electrons. The density of the core of the Sun is `10^(5) kg m^(–3)` . Assume that molar mass of proton is `1 "g mol"^(–1)` and the mass of an electron to be negligible compared to the mass of a proton. Estimate the pressure at the core of the sun. Assume that plasma behaves like an ideal gas. |
|
Answer» Correct Answer - `2xx10^(16) P_(a)` |
|
| 674. |
A Polyatomic gas with six degrees of freedom does 25 J of work when it is expanded at constant pressure The heat given to the gas isA. 100JB. 150JC. 200JD. 250J |
|
Answer» Correct Answer - A F = 6, W = 25 j `triangleQ = nc_ptriangleT=((6+2)/(2))nR triangle T` `triangleW = PtriangleV =nRtriangleT = 25` `triangleQ=4xx25=100` |
|
| 675. |
A gas is at 1 atm pressure with a volume `800 cm^(3)`. When `100 J` of heat is supplied to the gas, it expands to `1L` at constant pressure. The change in its internal energy isA. `80 J`B. `-80 J`C. `20 J`D. `-20 J` |
|
Answer» Correct Answer - A Work done by the gas, at constant pressure `DeltaW = PDeltaV` `=(1xx10^(5)Nm^(-2))(1000-800)xx10^(-6)m^(3)=20J` `DeltaU = DeltaQ-DeltaW`, so, `DeltaU=100J-20J = 80J`. |
|
| 676. |
A molecule of a gas has six degrees of freedom Then , the molar specific heat of the gas at constant volume isA. `(R)/(2)`B. RC. `(3R)/(2)`D. 3R |
|
Answer» Correct Answer - D We know that , specific heat of a gas (At Constant volume) `c_(V)=(f)/(2)xxR` where , R= number of independent relations f=degree of freedom `therefore " " C_(V)=(6)/(2)xxR=3R` |
|
| 677. |
A gas is at 1 atm pressure with a volume `800 cm^(3)`. When `100 J` of heat is supplied to the gas, it expands to `1L` at constant pressure. The change in its internal energy isA. `80 J`B. `- 80 J`C. `20 J`D. `- 20 J` |
|
Answer» Correct Answer - A a. Work done by the gas at constant pressure `Delta W = P Delta V` ` =(1 xx 10^(5) Nm^(-2)) (100 - 800) xx 10^(-6) m^(3) = 20 J` `Delta U = Delta Q - Delta W` `Delta U = 100 J - 20 J = 80 J` |
|
| 678. |
Find the number of degrees of freedom of molecules in a gas whose molar heat capacity at constant pressure is equal to `C_(P)=20J`/(moL K)A. 3B. 4C. 5D. 6 |
|
Answer» Correct Answer - A `C_(P)=(gammaR)/(gamma-1),gamma=(1+(2)/(f))` |
|
| 679. |
A reversible heat engine carries 1 mole of an ideal monatomic gas aroung the cycle 1-2-3-1. Process 1-2 takes place at constant volume, process 2-3 take place at constant volume, process 2-3 is adiabatic, and process 3-1 takes place at constant pressure. Complete the values for the heat `Delta Q`, the change in internal energy `Delta U`, and the work done `Delta `, for each of the three processes and for the cycle as a whole. |
|
Answer» For the process 1-2, we have `Delta W = 0` (since volume ramains constant) `Delta U = C, Delta T = (R )/(gamma - 1) Delta T = (8.3)/(((3)/(3))-1) xx (600 - 300) = 3735 J` For the process 2-3 we have `Delta Q = 0` (since the process is adiabatic) `Delta W = (p_(1) V_(1) - p_(2) V_(2))/(gamma - 1) = (R (T_(1) - T_(2)))/((gamma - 1)) = (8.3 (600 - 455))/(((5)/(3))- 1) = 1895 J` `Delta U = Delta Q - Delta W = 0 - 1805 = - 1805 J` For the process 3-1, we have `Delta Q = C_(P) Delta T = (gamma R)/((gamma - 1)) xx Delta T` or `Delta Q = (((5)/(3)) xx 8.3)/(((5)/(3))) xx (300 - 455) = - 3216 J` `Delta U = C_(V) Delta T = (R )/((gamma - 1)) Delta T = (8.3)/(((5)/(3)) -1) (300 - 455) = - 1286 J` `Delta W = Delta Q - Delta U = - 3116 - (- 1930) = - 1286 J` |
|
| 680. |
How many degress of freedom have the gas molecules, if under standard conditions the gas density is `rho = 1.3 kg//m^3` and velocity of sound propagation o it is `v = 330m//s` ? |
|
Answer» `v = sqrt((gamma P)/(rho)) implies gamma (v^(2) rho)/(P)` If `r` = degrees of freedom of rotational motion, them `gamma + 1 (2)/(r )` `:.r = (2)/((v^(2) rho)/(p) -1) = (2)/((330^(2) xx 1.3)/(1.015 xx 10^(5) - 1)) = 5` |
|
| 681. |
If `R = ` universal gas constant, the amount of heat needed to raise the temperature the temperature of `2 mol` of an ideal monatomic gas from `273 K` to `373 K` when no work is done isA. `100 R`B. `150 R`C. `300 R`D. `500 R` |
|
Answer» Correct Answer - C c. `Delta Q = Delta U` `= mu C_(v) Delta T = mu ((R )/(gamma - 1)) Delta T = 2 xx (R )/((5)/(3) -1) [373 - 273] = 300 R` (as for monatomic gas `gamma = 5//3`) |
|
| 682. |
A reversible heat engine carries `1 mol` of an ideal monatomic gas around the cycle `ABCA`, as shown in the diagram. The process `BC` is adiabatic. Call the processes `AB, BC` and `CA` as `1,2` and `3` and the heat `( DeltaQ)_(r)`, change in internal energy `(DeltaU)`, and work done `( DeltaW)_(r), r=1,2,3` respectively. The temperature at `A,B,C` are `T_(1)=300K,T_(2)=600K` and `T_(3)=455K`. Indicate the pressure and volume at `A,B` and` C` by `P_(r)` and `V_(r), r=1,2,3,` respectively. Assume that intially pressure `P_(1)=1.00atm.` Which of the following represents the correct values of the quantities indicated ? A. `(DeltaQ)=450RJ, (DeltaU)_(1)=450RJ,(DeltaW)_(1)=300RJ`B. `(Delta Q)_(2)=0, (DeltaU)_(2)=450RJ, (DeltaW)_(2)=-217.5 RJ`C. `(DeltaQ)_(3)=0, (DeltaU)_(3)=-232.5RJ,(DeltaW)_(3)=0`D. `(DeltaQ)_(1)=450RJ, (DeltaU)_(3)=-232.5RJ, (DeltaW)_(2)=217.5RJ` |
|
Answer» Correct Answer - D Process 1 is isochoric. Now work is done `Dw=0` Hence, `(DeltaQ)_(1)=dU=C_(V)(T_(2)-T_(1))` `=(3R)/(2)[600-300]=450RJ` Process 2 is adaibatic expansion `(Delta Q)_(2)=0,(DeltaW)_(2)=-(DeltaU)_(2)` `(DeltaW)_(2)=-[C_(v)(T_(3)-T_(2))]` `=-[(3R)/(2)(455-600)]` `=(3xx145)/(2)R=217.5RJ` `(DeltaV)_(3)=C_(v)-217.5RJ` Process 3 is isobaric compression. Value of pressure `= 1 atm = 1.013 xx 10^(5)N//m^(2)` `(DeltaW)_(3)=1.013xx10^(5)(V_(c)-V_(A))J=0` `(DeltaW)_(3)=C_(v)(300-455)=(-3R)/(2) xx 155=-232.5R` |
|
| 683. |
The average degrees of freedom per molecule for a gas are 6. The gas performs `25 J` of work when it expands at constant pressure. The heat absorbed by gas isA. `75 J`B. `100 J`C. `150 J`D. `125 J` |
|
Answer» Correct Answer - B b. As `f = 6` (given), therefore `gamma = 1 + (2)/(f) = 1 + (2)/(6) = (4)/(3)` Fraction of energy given for external work `(Delta W)/(Delta Q) = (1 - (1)/(gamma))` `implies (25)/(Delta Q) = (1 - (1)/(4//3)) = 1 - (3)/(4) = (1)/(4)` |
|
| 684. |
Find the molar mass and the number of degrees of freedom of molecules in a gas if its heat capacities are `C_(v) = 650 J kg^(-1) K^(-1)` and `C_(P) = 910 J kg^(-1) K^(-1)` |
|
Answer» `C_(V) = (R )/(gamma -1)` and `C_(V) = Mc_(v)` `:. M xx 650 = (8.3)/(gamma -1)` Futher `gamma C_(p) // C_(v) = (910)/(650)` Solving Eqs. (i) and (ii), `M = 32 g//mol` `gamma - 1,4`, so the gas is diatomic and the number of degrees of freedom is 5. |
|
| 685. |
If `50 cal` of heat is supplied to the system containing `2 mol` of an ideal monatomic gas, the rise in temperature is `(R = 2 cal//mol-K)`A. `50 K`B. `5 K`C. `10 K`D. `20 K` |
|
Answer» Correct Answer - B b. `(C_(v))/(C_(P)) xx Q = nC_(V) dT` `dT = (Q)/(nC_(P)) = (50)/(2 xx (5)/(2) xx R) = 5 K` |
|
| 686. |
Five moles of hydrogen gas are heated from `30^(@)C` to `60^(@)C` at constant pressure. Heat given to the gas is (given `R = 2 cal//mol` degrees)A. `750 cal`B. `630 cal`C. `1050 cal`D. `1470 cal` |
|
Answer» Correct Answer - C c. `(Delta Q)_(P) = mu C_(P) Delta T = mu ((gamma)/(gamma - 1)) R Delta T` `:. (Delta Q)_(P) = 5 xx (((7)/(5))/((7)/(5) - 1)) xx 2 xx 30` `= 5 xx 2 xx (7)/(5) xx (5)/(2) xx 30 = 1050 cal` (as `mu = 5` mol and `gamma = 7//5` for `H_(2))` |
|
| 687. |
If the specific heat of lead is 0.03 cal/gm, then the thermal capacity of 500 gm of lead will beA. `5cal//.^(@)C`B. `10cal//.^(@)C`C. `15cal//.^(@)C`D. `20cal//.^(@)C` |
|
Answer» Correct Answer - C `C=mc=500xx0.03=15cal//.^(@)C` |
|
| 688. |
The molar specific heat of oxygen at constant pressure `C_(P) = 7.03 cal//mol .^(@)C` and `R = 8.31 J//mol .^(@)C`. The amount of heat taken by 5 mol of oxygen when heated at constant volume from `10^(@)C` to `20^(@)C` will be approximately.A. 25 calB. 50 calC. 250 calD. 500 cal |
|
Answer» Correct Answer - C c. `Q = nC_(v) Delta T = n (C_(P) - R) Delta T` `= 5 (7.03 - (8.31)/(4.2)) xx (20 - 10) = 250 cal` |
|
| 689. |
Molar specific heat of oxygen at constant pressure is 7.2 cal/mol `.^(@)C` and R=8.3 J/mol K. at constant volume, 5 moles of oxygen is heated from `10^(@)C` to `20^(@)C` then the quantity of heat required will beA. 25 calB. 50 calC. 260 calD. 500 cal |
|
Answer» Correct Answer - C `C_(V)=C_(p)-R` `=7.2-2=5.2` `dQ_(V)=nC_(V)dT` `=5xx5.2xx10=260cal`. |
|
| 690. |
The molar specific heat of oxygen at constant pressure `C_(P) = 7.03 cal//mol .^(@)C` and `R = 8.31 J//mol .^(@)C`. The amount of heat taken by 5 mol of oxygen when heated at constant volume from `10^(@)C` to `20^(@)C` will be approximately.A. 25 calB. 50 calC. 250.5 calD. 500 cal |
|
Answer» Correct Answer - C `dQ_(V)=nC_(V)dT` `=5xx(C_(p)-R)dT` `=5(7.03-2)xx10=5xx5xx10` =250 cal |
|
| 691. |
The change in internal energy when 5 mole of hydrogen is heated to `20^(@)C` from `10^(@)C`, specific heat of hydrogen at constant pressure is 8 cal/mol`.^(@)C` isA. 200 calB. 350 calC. 300 calD. 475 cal |
|
Answer» Correct Answer - C `C_(V)=C_(p)-R=8-2=6cal//mol.^(@)C` du=n`C_(V)(T_(2)T_(1))=5xx6xx10=300` |
|
| 692. |
One mole of mono atomic gas is mixed with 2 moles of diatomic gas. What is the value of molecular specific heat of the gas at constant prssure in cal/mol K? (R=2 cal/mol K)A. 1.5B. 6.32C. 3.5D. 4.16 |
|
Answer» Correct Answer - B `C_("p mix")=(n_(1)C_(p_(1))+n_(2)C_(p_(2)))/(n_(1)+n_(2))` |
|
| 693. |
Total random kinetic energy of the molecules in 1 mole of a gas at a temperature of 300 K is (R = 2 cal/mole °C)A. 900 calB. 450 calC. 2250 calD. 600 cal |
|
Answer» Correct Answer - A `K.E=(3)/(2)RT=(3)/(2)xx2xx300=900cal`. |
|
| 694. |
What is the rms speed of oxygen molecules at `225^(@)C` ? Density of oxygen at NTP is `1.42 kg m^(-3)` and one atmosphere is `1.013xx10^(5)Nm^(-2)` .A. `624.8ms^(-1)`B. `618.6ms^(-1)`C. `328.5ms^(-1)`D. `320.7ms^(-1)` |
|
Answer» Correct Answer - A Given , temparature, T=225+273=498K, density,`p=1.42" kg m"^(-3)` pressure , `p=1.013xx10^(5)Nm^(-2)` and `rms speed , v_(rms)` ? The rms speed of oxygen molecules at NTP, `v_(rms)=sqrt((3p)/(p))=sqrt((3xx1.013xx10^(5))/(1.42))=4.626xx10^(2)` `v_(rms)=462.6ms^(-1)` Now , rms speed at 489 K `("or" 225^(@)C)` `because v_(rms)propsqrt(T)` `therefore ((v_("rms")))/((v_("rms"))_(O_(2)))=sqrt((T)/(T_(0)))rArr((v_("rms")))/(462.6)=sqrt((498)/(273))` `v_("rms")=1.350xx462.6=624.80ms^(-1)` |
|
| 695. |
The mean kinetic energy of one gram-mole of a perfect gas at abolute temperature T isA. `(1)/(2)kT`B. `(1)/(2)RT`C. `(3)/(2)kT`D. `(3)/(2)RT` |
|
Answer» Correct Answer - D `KE=(3)/(2)RT`. |
|
| 696. |
If at NTP, velocity of sound in a gas is 1150 m/s, then find out the rms velocity of gas molecules at NTP. (Given R=8.3J/mol/K, `C_(P)=4.8` cal/mol/k).A. `1600"ms"^(-1)`B. `1532.19"ms"^(-1)`C. `160"ms"^(-1)`D. Zero |
|
Answer» Correct Answer - B `R= (8.3)/(4.2)"cal "g"mol"^(-1)-K^(-1)` `C_(v)=C_(p)-R=(4.8-(8.3)/(4.2))=2.824` `gamma=(C_(p))/(C_(v))=(4.8)/(2.824)=1.69` `"Since", " " v=sqrt(((3)/(gamma)))v_(s)=sqrt((3)/(1.69))xx1150=1532.19"ms"^(-1)` |
|
| 697. |
At what temperature will the average KE per molecule of a gas be exactly half its value at NTP?A. 150 KB. 136.5 KC. 273 KD. 546 K |
|
Answer» Correct Answer - B `E_(2)=(E_(1))/(2) and T_(1)=273` `(E_(2))/(E_(1))=(T_(2))/(T_(1))` `therefore T_(2)=(E_(2))/(E_(1))` `T_(1)=(273)/(2)` `therefore T_(2)=136.5K` |
|
| 698. |
PV/3=RT, V represents volume ofA. any amount of gasB. 2 moles of gasC. 3 moles of gasD. 4 moles of gas |
|
Answer» Correct Answer - C `(PV)/(3)=RT` `PV=3RT` n=3 moles. |
|
| 699. |
For a gram molecules of a gas, the value of the constant R in the equation PV=RT will beA. 2 cal/mol KB. 0.4 cal/mol KC. 8 cal/mol KD. 8.3 cal/mol K |
|
Answer» Correct Answer - A Universal gas constant have value R=2 cal/mol `.^(@)K`. |
|
| 700. |
A sound wave passing through air at `NTP` produces a pressure of `0.001 dyne//cm^(2)` during a compression. The corresponding change in temperature (given `gamma = 1.5` and assume gas to be ideal) isA. `8.97 xx 10^(-4) K`B. `8.97 xx 10^(-6) K`C. `8.97 xx 10^(-8) K`D. None of these |
|
Answer» Correct Answer - C Differentiating `gammaT^(gamma-1) dT P_(1-gamma)+T^(gamma)(1-gamma)P^(-gamma)dP = 0` or `dT = ((gamma-1)T)/(gamma P) dP` `or dT = ((1.5-1)/(1.5))((273)/(76xx13.6xx981)xx0.001)` `=8.97xx10^(-8)K`. |
|