871 + Interview Questions in KINETIC-THEORY-OF-GASES IN GENERAL AWARENESS Page 11 InterviewSolution

501.	`1 g` mole of an ideal gas at STP is subjected to a reversible adiabatic expansion to double its volume. Find the change in internal energy `( gamma = 1.4)`A. `1169.5 J`B. `769.5 J`C. `1369.5 J`D. `969.5 J`
Answer» Correct Answer - C Use `T_(1)V_(1)^(gamma-1) = T_(2)^(gamma-1)` `or T_(2) = (T_(1)V_(1)^(gamma-1))/(V_(2)^(gamma-1)) = (273)/((2)^(0.4)) = 207 K`. Change in internal energy `Delta U = (R)/((gamma-1)) (T_(1)-T_(2)) = (8.31(273-207))/(1.4-1) = 1369.5J`.

Discussion

502.	Heat given to a system is 35 joules and work done by the system is 15 joules. The change in the internal energy of the system will beA. `-50J`B. `20J`C. `30J`D. 50J
Answer» Correct Answer - B `dQ=dU+dWrArrdW=dQ-dU`

Discussion

503.	Find the chagne in internal energy in joule when 10g of air is heated from `30^(@)C` to `40^(@)C` `(c_(V)=0.172 kcal//kg//kj=4200j//kcal)`A. `62.24J`B. `72.24J`C. `52.24J`D. `82.24J`
Answer» Correct Answer - B `W=(P(V_(2)-V_(1))`

Discussion

504.	The relation between the internal energy `U` adiabatic constant `gamma` isA. `U = (PV)/(gamma -1)`B. `U = (PV^(gamma))/(gamma -1)`C. `U = (PV)/(gamma)`D. `U = (gamma)/(PV)`
Answer» Correct Answer - A a. Change in internal energy `Delta U = mu C_(v) Delta T implies U_(2) - U_(1) = mu C_(v) (T_(2) - T_(1))` Let initially `T_(1) = 0` so `U_(1) = 0` and finally `T_(2) = T` and `U_(2) = U` `U = mu C_(v) T = mu T xx C_(v) = (PV)/(R ) xx (R )/(gamma - 1) = (PV)/(gamma - 1)` (As `PV = mu RT, :. mu T = PV//R` and `C_(v) = R//(gamma - 1))`

Discussion

505.	The temperature and the dew point in an open room are `20^(@)C and 10^(@)C`.if the room temperature drops to `15^(@)C`,what will be the new dew point?
Answer» Temperature = ` 20^@C,` dew point = ` 10^@C` The place is saturated at ` 10^@C` even if the temperature drop dew point remains unaffected The air has the vapour pressure which is the saturation VP at ` 10^@C` if saturated vapour pressure does not change on temperature .

Discussion

506.	There is some liquid in a closed bottle. The amount of liquid is continuously decreasing.The vapour in the remaining partA. (a)must be saturatedB. (b)must be unsaturatedC. (c)may be unsaturatedD. (d)there will be no vapour.
Answer» Correct Answer - B

Discussion

507.	To what temperature should the hydrogen at `327^(@)C` be cooled at constant pressure, so that the root mean square velocity of its molecules become half of its previous value?A. `-123^(@)C`B. `23^(@)C`C. `-100^(@)C`D. `0^(@)`
Answer» Correct Answer - A `v_(rms)=sqrt((RT)/(M))rArrv_(rms)propsqrt(T)`

Discussion

508.	A gas expands adiabatically at constant pressure such that its temperature `Tprop(1)/(sqrt(V))` , the value of `C_(P)//C_(V)` of gas isA. `1.30`B. `1.50`C. `1.67`D. `2.00`
Answer» Correct Answer - B For adiabatic expansion , we have the formula `pV^(gamma)` = Constant ... (i) Gas equation is , `pV=RTrArrp=(RT)/(V)` …(ii) From Eqs . (i) and (ii) , we obtain `((RT)/(V))V^(gamma)` = constant `rArr" " TV^(gamma-1)` = constant ...(iii) `"But " Tprop(1)/(sqrtV)` (given) ` "as " TV^(1//2)` = constant ...(iv) THus , using Eqs. (iii)and (iv) together, we get `gamma-1=(1)/(2)` `"or " gamma=(3)/(2)=1.5rArr(C_(p))/(C_(v))=1.5`

Discussion

509.	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.
Answer» Correct Answer - A

Discussion

510.	For adiabatic process of an ideal gas the value of `(dP)/P` is equals toA. `-gamma (dV)/V`B. `-gamma V/(dV)`C. `(dV)/V`D. `-gamma^(2) (dV)/V`
Answer» Correct Answer - A `(dP)/P=-gamma(dV)/V`

Discussion

511.	A Carnot engine used first ideal monoatomic gas and then an ideal diatomic gas , if the source and sink temperatures are `411^(@)C` and `69^(@)C` ,respectively and the engine extracts 1000 J of heat from the source in each cycle , thenA. area enclosed by the p-V diagram is 10 JB. heat energy rejected by engine is 1 st case is 600 J while that in 2 nd case in 113 JC. area enclosed by the p-V diagram is 500 JD. efficiencies of the engine in both the cases are in the ratio 21 : 25
Answer» Correct Answer - C Here , `T_(1)=411^(@)C=(411+273)K=684K` `T_(2)=69^(@)C=(69+273)K=342Kand Q_(1)=1000J` `because"Efficiency "eta=(W)/(Q_(1))=1-(T_(2))/(T_(1))=1-(342)/(684)=(1)/(2)` `rArr" " W=etaQ_(1)=(1000)/(2)=500J`

Discussion

512.	The coeffcient of performance of a refrigerator working between `10^(@)C` and `20^(@)C` isA. 28.3B. 29.3C. 2D. Cannot be calculated
Answer» Correct Answer - A Given , `T_(1)=273+20=293K,T_(2)=273+10=283K` `therefore` Conefficient performance = `(T_(2))/((T_(1)-T_(2)))=(283)/(293-283)=(283)/(10)=28.3`

Discussion

513.	On a cold winter night you switch on a room heater to keep your room warm. Does it mean that the total energy of the air inside the room increases after you switch on the heater?
Answer» Correct Answer - No

Discussion

514.	Consider a collision between an oxygen molecule and a hydrogen moleule in a mixture of oxygen and hydrogen kept at room temperature. Which of the following aer possible?A. (a)The kinetic energies of both the molecules increase.B. (b)the kinetic energies of both the molecules decrease.C. (c)Kinetic energy of the oxygen molecule increases and that of the hydrogen molecule decreases.D. (d)The kinetic energy of the hydrogen molecule increases and that of the oxygen molecule decreases.
Answer» Correct Answer - C::D

Discussion

515.	Heat is supplied to a certain homogeneous sample of matter, at a uniform rate. Its temperature is plotted against time, as shown Which of the following conclusions can be drawn? (i) Its specific heat capacity is greater in the solid state than the liquid state. (ii) Its specific heat capacity is greater in the liquid state than in the solid state. (iii) Its latent heat of vaporization is greater than its latent heat of fusion. (iv) Its latent heat of vaporization is smaller than its latent heat of fusionA. Its specific heat capacity is equal in the solid state than in the liquid state.B. Its specific heat capacity is greater in the liquid state than in the solid state.C. its talent heat vaporization is greater than its latent heat of fusion.D. its latent heat of vaporization is smaller than its latent heat of fusion.
Answer» Correct Answer - C Slope of graph is greater in the solid state i.e., temperature is rising faster, hence lower heat capacity The transition from solid to liquid state takes lesser time, hence latent heat is smaller.

Discussion

516.	The root mean square velocity of the molecules in a sample of helium is `5//7th` that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is `0^(@)C`, then the temperature of the helium sample is aboutA. `0^(@)C`B. 4 KC. `273^(@)C`D. `100^(@)C`
Answer» Correct Answer - C As, `(c_(He)/c_(H))=sqrt((gammarho)/(rho))=sqrt((1)/(4))=(1)/(2)" or" (c_(He))t=(c_(He))_(0)sqrt((T)/(T_(0)))` `therefore" " (c_(He))_(t)/(c_(H))_(0)=(c_(He))_(0)/(c_(H))_(0)sqrt((T)/(t))=(5)/(7)` `"or" " " T~~2T_(0)=2xx273=546^(@)K=273^(@)C`

Discussion

517.	In the cylinder shown in the figure, air is enclosed under the piston. Piston mass M=60 kg, crosssectional area of the cylinder `S_(0)=20cm^(2)` atmospheric pressure `P_(0)=10^(5) pa alpha=37^(@)`. The air temperature is constant, the friction is negligible. What is the pressure of the enclosed gas?A. `3.4 xx 10^5 Pa`B. `4 xx 10^5 Pa`C. `4.75 xx 10^5 Pa`D. `2.8 xx 10^5 Pa`
Answer» Correct Answer - B `P_0 xx S_0 + mg = P xx S_0 sec 37^(@ )xx cos 37^(@)` `rArr P = P_0 + (msg)/(S_0)` `rArr 10^5 + (600)/(20 xx 10 ^-4)` `=4 xx 10^5 Pa`

Discussion

518.	The temperature and the relative humidity are `300 K and 20%` in a room of volume `50 m^(3)`. The floor is washed with water, `500 g` of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at `300 K = 3.3 kPa`.
Answer» `RH = (VP/SVP)` ` rArr 0.20 = VP/3.3 xx (10^3)` ` rArr VP = 0.20 xx 3.3 xx (10^3)` = 660 rArr PV = nRT ` rArr P = nRT/V = m/M xx (R xx T/V)` ` = (500 xx 8.3 xx 300/ 18 xx 50)` ` = 1383.3` Net P = 1383.3 + 660 = 2043.3 Now , RH = 2043.3/3300 = 0.619 = 62%.

Discussion

519.	An engine has an efficiency of `1/6`. When the temperature of sink is reduced by `62^(@)C`, its efficiency is doubled. Temperature of the source isA. `127^(@)C`B. `37^(@)C`C. `62^(@)C`D. `99^(@)C`
Answer» Correct Answer - D Effeciency of engine is given by , `eta = 1-(T_2)/(T_1)` `:. (T_2)/(T_1) = 1-eta=1-1/6=5/6` ..(i) In other case `(T_(2)-62)/(T_1) = 1-eta=1-2/6=2/3` ..(ii) Using eq(i) `T_(2)-62=2/3T_(1)=2/3xx6/5 T_(2) or 1/5 T_(2)=62` `T_(2) = 310K = 310-273^@C=37^@C` Hence, `T_(1) = 6/5 T_(2)=6/5xx310=372K` Temperature at `.^(@)C=372-273=99^(@)C` Hence, temperature of sourve is `99^(@)C`.

Discussion

520.	In the above problem work done along path BC is:A. zeroB. (40 - 20 )= 20 JouleC. 40 JouleD. 60 Joule
Answer» Correct Answer - A `W_(BC) = 0 as triangleV =0`

Discussion

521.	The air density at mount Everest is less than that at the sea level . It is found by mountainers that for one trip lasting few hours , the extra oxygen needed by them corresponds to 30000 cc at sea level (pressure 1 atm , temperature `27^(@)C` ). Assuming that the temperature around Mount Everested is `-73^(@)C` and that the pressure at which oxygen be filled (at site) in the cylinder isA. 3.86 atmB. 5.00 atmC. 5.77atmD. 1 atm
Answer» Correct Answer - A Here, `p_(1)=1"atm",T_(1)=300K.V_(1)=30000cc=30L` `and" " T_(2)=200K,V_(2)=5.2L` `therefore" " p_(2)=(p_(1)V_(1)T_(2))/(T_(1)V_(1))=(1xx30xx200)/(300xx5.2)=3.86"atm`

Discussion

522.	A kettle with 2 litre water at `27^@C` is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate `160 J//s`, when its lid is open. In how much time will water heated to `77^@C` with the lid open ? (specific heat of water = `4.2 kJ//^@C.kg)`A. 5 min 40 sB. 10 min 20 sC. 8 min 20 sD. 16 min 10 s
Answer» Correct Answer - C Rate of heat gain =`1000-160=840J//s` `therefore` Required time = `(2xx4.2xx10^(3)xx(77-27))/(840)=500s` = 8 min 20 s

Discussion

523.	A planet is at an average distance d from the sun and its average surface temeperature is T. Assume that the planet receives energy only from the sun and loses energy only through radiation from the surface. Neglect atmospheric effects. If `Tpropd^(-n)`, the value of n isA. 2B. 1C. `1//2`D. `1//4`
Answer» Correct Answer - C Let R = radius of planet P=powerradiated by the sun Energy received by planet =`(p)/(4pid^(2))xx4piR^(2)` Energy radiated by planet = `(4piR)^(2)sigmaT^(4)` For thermal equilibrium , `4piR^(2)(sigmaT^(4))=(P)/(4pid^(2))xx4piR^(2)` `T^(4)prop(1)/(d^(2))rArrTpropd^(-1//2)" " [thereforen=(1)/(2)]`

Discussion

524.	The heat engine would operate by taking heat at a particular temperature andA. converting it all into workB. converting some of it into work and rejecting the rest at lower temperatureC. converting some of it into work and rejecting the rest at same temperatureD. converting some of it into work and rejecting the rest at a higher temperature.
Answer» Correct Answer - B

Discussion

525.	Consider a heat engine as shown in (figure). `Q_(1) and Q_(2)` are heat added to heat bath `T_(1)` and heat taken from `T_(2)` one cycle of engine. `W` is the mechanical work done on the engine. If `Wgt0`, then possibillities are:A. `Q_(1)gtQ_(2)gt0`B. `Q_(2)gtQ_(1)gt0`C. `Q_(3)ltQ_(1)lt0`D. `Q_(1)lt0,Q_(2)gt0`
Answer» Correct Answer - A::C Consider the figure we can write `Q_(1)=W+Q_(2)` `rArrW=Q_(1)gtQ_(2)gt0` (if both `Q_(1)` and `Q_(2)` are positive) We can also, write `Q_(2)ltQ_(1)lt0` (if both `Q_(1)` and `Q_(2)` are negative).

Discussion

526.	Find the external work done by the system in keal, when 20 kcal of heat is supplied to the system and the increase in the internal energy is 8400J (J=4200 J//kcal)?A. 16kcalB. 18kcalC. 20kcalD. 19 kcal
Answer» Correct Answer - B `mgh=JH`

Discussion

527.	Calculate the kinetic energy of translation of the molecules of `20g` of `CO_(2)` at `27^(@)C`.
Answer» Correct Answer - `1697.7J`

Discussion

528.	If three gas molecules have velocity `0.5`, `1` and `2 km//s` respectively, find the ratio of their root mean square speed and average speed.A. `V_(rms)=(V_(avg))/(2)`B. `V_(rms)=V_(avg)`C. `V(r.m.s)=1.134V_(av)`D. `V_(rms)=2V_(avg)`
Answer» Correct Answer - C `V_(rms)=sqrt((V_(1)^(2)+V_(2)^(2)+V_(3)^(2))/(3))` =1.323 `V_(rms)=(V_(1)+V_(2)+V_(3))/(3)=1.67km//s` `V_(rms)=1.134V_(avg)`

Discussion

529.	Consider a gas at tmperature T occupying a volume V consisting of a mixture of two gases having. `N_a & N_b` atoms of masses `m_a & m_b` respsectively. Give an expression for the total pressure exerted by the gasA. `((N_a + N_b)kT)/(V)`B. `(1.5(N_a + N_b)kT)/(V)`C. `((N_a +N_b)RT)/(V)`D. none
Answer» Correct Answer - A ` P = (nRT)/(V)`, where n `=` number of moles of all gases `therefore n = (N_a + N_b)/(N)`, N = Avogadro number `P= ((N_a+N_b))/(N)(RT)/(V) = (N_a + N_b)/(V)(R)/(N)T=((N_a+N_b)kT)/(V)`

Discussion

530.	Derive an expression for pressure exterted by an ideal gas?A. `((N_(a)+N_(b))kT)/(V)`B. `(1.5(N_(a)+N_(b))kT)/(V)`C. `((N_(a)+N_(b))RT)/(V)`D. none
Answer» Correct Answer - C After getting combined the no of moles becomes half in this case `because` at const.V and T `P propto n` ` `therefore` P beacomes half

Discussion

531.	Assertion : The molecules of a monatomic gas has three degrees freedom. Reason : The molecules of a diatomic gas has five degrees of freedom.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
Answer» Correct Answer - B A monatomic gas molecules consists of a single atom. It can have translational motion in any direction in space. Thus, it has 3 translation degree of freedom. `f = 3 ` all transnation it can also rotate but due to its small moment of inertial rotational kinetic energy is neglected. The molecules of a diatomic gas(like `O_(2),CO_(2),H_(2))` cannot only move bodily but also rotate about any one of the three co-ordinate axes. Hence, it can have only two rotational degree of freedom. Thus a diatomic molecule has 5 degree of freedom. 3 translational and 2 rotational.

Discussion

532.	A gas take part in two thermal processes in which it is heated from the same initial state to the same final temperature. The processes are shown on the `P-V` diagram by straight lines `1 rarr 3 ` and `1 rarr 2` A. `W_(13)gt W_(12)`B. `W_(12)gt W_(13)`C. `W_(13)= W_(12)`D. None of above
Answer» Correct Answer - A Since the temperature remains constant, `Delta T=0` Now `W_(12)=(1)/(2) (P_(1)+P_(2))(V_(2)-V_(1))` `W_(13)=(1)/(2) ( P_(1)+P_(3))(V_(3)-V_(1))` `W_(12)-W_(13)=(1)/(2)[P_(1)(V_(2)-V_(3))+(P_(3)-P_(1))]lt0` `W_(13)gtW_(12), Q_(13)gtQ_(12)`

Discussion

533.	A gas take part in two thermal processes in which it is heated from the same initial state to the same final temperature. The processes are shown on the `P-V` diagram by straight lines `1 rarr 3 ` and `1 rarr 2` A. `Q_(13) lt Q_(12)`B. `Q_(13) gt Q_(12)`C. `Q_(13) = Q_(12)`D. None of above
Answer» Correct Answer - B Since the temperature remains constant, `Delta T=0` Now `W_(12)=(1)/(2) (P_(1)+P_(2))(V_(2)-V_(1))` `W_(13)=(1)/(2) ( P_(1)+P_(3))(V_(3)-V_(1))` `W_(12)-W_(13)=(1)/(2)[P_(1)(V_(2)-V_(3))+(P_(3)-P_(1))]lt0` `W_(13)gtW_(12), Q_(13)gtQ_(12)`

Discussion

534.	Two temperature scales A and B are related by : `(A-42)/(110)=(B-72)/(220)` At which temperature two scales have the same reading ?A. `-42^(@)`B. `-72^(@)`C. `+12^(@)`D. `-40^(@)`
Answer» Correct Answer - C We are given the reation between the tempeature scales A and B as `(A-42)/(110)=(B-72)/(220)` , so for the two scales to have the same reading A = B and so `(A-42)/(110)=(A-72)/(220)` which yields A=12.

Discussion

535.	We know that volume (V ) occupied by a substance scales as `x^(3)` , where x is the average distance between its molecules. Assume that water vapour at `100^(@)C` and atmospheric pressure has average intermolecular separation equal to `x_(v)` and for liquid water at `100^(@)C` the intermolecular separation is `x_(w)`. Find the ratio `(X_(v))/(x_(w))` considering density of water at `100^(@)C` to be `1.0 x× 103 kg//m^(3)` and taking water vapour as an ideal gas. Atmospheric pressure `P_(0) = 1.0 x× 10^(5) N//m^(2)` , Gas constant `R = 8.3 "J mol"^(–1) K^(–1)`.
Answer» Correct Answer - 12

Discussion

536.	The work done on the system in an aidabatic compression depends onA. the increase in internal energy of the systemB. the decrease in internal energyC. the change in volume of the systemD. all the above
Answer» Correct Answer - A

Discussion

537.	Suppose there are N molecules each of mass m, of an ideal gas in a container. The x component of velocity of a molecule is denoted by `u_(x)`. The gas is enclosed using a horizontal piston of area A as shown. If the piston is moved in so as to reduce the volume of gas by half, keeping the temperature of gas constant, we know from the gas law that the pressure will be doubled. On microscopic level the increase in pressure on the Vertical side walls is becauseA. momentum change per collision is doubled while the frequency of collision remains constantB. momentum change per collision remain constant while the frequency of collision is doubledC. momentum change per collision and the frequency of collision both are increasedD. none of these two physical quantities are changed. It is due to some other reason.
Answer» Correct Answer - D

Discussion

538.	Suppose there are N molecules each of mass m, of an ideal gas in a container. The x component of velocity of a molecule is denoted by `u_(x)`. The gas is enclosed using a horizontal piston of area A as shown. The pressure of the gas can also be written in terms of momentum transferred per collision `(Deltap)` and collision frequency (f) on the wall of area A asA. `Deltapxxf`B. `(Deltap)/(Axxf)`C. `(Deltapxxf)/(A)`D. none of these
Answer» Correct Answer - C

Discussion

539.	Suppose there are N molecules each of mass m, of an ideal gas in a container. The x component of velocity of a molecule is denoted by `u_(x)`. The gas is enclosed using a horizontal piston of area A as shown. The pressure of the gas isA. `(m(u_(x)^(2))N)/(AL)`B. `(m(u_(x)^(2))N)/(3AL)`C. `(3m(u_(x)^(2))N)/(AL)`D. `(3m(u_(x)^(2)))/(AL)`
Answer» Correct Answer - A

Discussion

540.	During adiabatic compression of a gas, its temperatureA. fallsB. raisesC. remains constantD. becomes zero
Answer» Correct Answer - B

Discussion

541.	In the adiabatic compression the decrease in volume is associated withA. increase in temperature and decrease in pressureB. decrease in temperature and increase in pressureC. decreae in temperature and decrease in pressureD. increase in temperature and increae in pressure
Answer» Correct Answer - D

Discussion

542.	For adiabatic processes `(gamma = (C_(p))/(C_(v)))`A. `P^(1-gamma)=` constantB. `P^(gamma)T^(1-gamma)=` constantC. `PT^(gamma)=` cosntantD. `P^(gamma)T=` constant
Answer» Correct Answer - A

Discussion

543.	In an adiabatic process, `R = (2)/(3) C_(v)`. The pressure of the gas will be proportional to:A. `T^(5//3)`B. `T^(5//2)`C. `T^(5//4)`D. `T^(5//6)`
Answer» Correct Answer - B `R = (2)/(3) C_(v)` We know, `C_(P) - C_(V ) = R` or `gamma - 1 = (R )/(C_(V))` or `R = C_(V) (gamma - 1)` Comparing `gamma = 5//3` `P^(1 - gamma) T^(gamma) =` constant = `K` or `P prop T^((gamma//gamma -))` Given `gamma = 5//3` `gamma//(gamma - 1) = 5//2` so, `P prop T^(5//2)`

Discussion

544.	If an ideal gas is isothermally expanded its internal energy willA. increaseB. decreaseC. remains sameD. decrease or increase depending on nature of the gas
Answer» Correct Answer - C

Discussion

545.	The temperature of the system decreases in the process ofA. free expansionB. isothermal expansionC. adiabatic expansionD. isothermal compression
Answer» Correct Answer - C

Discussion

546.	In a adiabatic process is increased by `2//3%` if `C_(P)//C_(V) = 3//2`. Then the volume decreases by aboutA. `(4)/(9)%`B. `(2)/(3)%`C. `4%`D. `(9)/(4)%`
Answer» Correct Answer - A a. For a adiabatic process, `Pv^(gamma) = K` Here, `gamma = 3//2` and `K =` constant `:. PV^(3//2) = K` `log P + (3)/(2) log V = log K` `(Delta P)/(P) + (3)/(2) (Delta V)/(V) = 0` `:. (Delta V)/(V) = (2)/(3) = (Delta P)/(P)` `:. (Delta V)/(V) xx 100 = - ((2)/(3)) ((Delta P)/(P) xx 100) = - (2)/(3) xx (2)/(3) = - (4)/(9)` Therefore volume decrease decrease by about `(4//9) %`.

Discussion

547.	For an adiabatic process the relation between V and T is given byA. `TV^(gamma)=` constantB. `T^(gamma)V=`constantC. `TV^(1-gamma)=` constantD. `TV^(gamma-1)=`constant
Answer» Correct Answer - D

Discussion

548.	The temperature of the system decreases in the process ofA. free expansionB. adiabatic expansionC. isothermal expansionD. isothermal compression
Answer» Correct Answer - B

Discussion

549.	Heat engine rejects some heat to the sink. This heatA. converts into electrical energy.B. converts into light energy.C. converts into electromagnetic energyD. is unavailable in the universe.
Answer» Correct Answer - D

Discussion

550.	For an adiabatic change in a gas, if P,V,T denotes pressure, volume and absolute temperature of a gas at any time and `gamma` is the ratio of specific heats of the gas, then which of the following equation is true?A. `T^(gamma)P^(1-gamma)=` const.B. `T^(1-gamma)P^(gamma)=` const.C. `T^(gamma-1)V^(gamma)=` const.D. `T^(gamma)V^(gamma)=` const.
Answer» Correct Answer - A

Discussion

Explore topic-wise InterviewSolutions in .

`1 g` mole of an ideal gas at STP is subjected to a reversible adiabatic expansion to double its volume. Find the change in internal energy `( gamma = 1.4)`A. `1169.5 J`B. `769.5 J`C. `1369.5 J`D. `969.5 J`

Heat given to a system is 35 joules and work done by the system is 15 joules. The change in the internal energy of the system will beA. `-50J`B. `20J`C. `30J`D. 50J

Find the chagne in internal energy in joule when 10g of air is heated from `30^(@)C` to `40^(@)C` `(c_(V)=0.172 kcal//kg//kj=4200j//kcal)`A. `62.24J`B. `72.24J`C. `52.24J`D. `82.24J`

The relation between the internal energy `U` adiabatic constant `gamma` isA. `U = (PV)/(gamma -1)`B. `U = (PV^(gamma))/(gamma -1)`C. `U = (PV)/(gamma)`D. `U = (gamma)/(PV)`

The temperature and the dew point in an open room are `20^(@)C and 10^(@)C`.if the room temperature drops to `15^(@)C`,what will be the new dew point?

There is some liquid in a closed bottle. The amount of liquid is continuously decreasing.The vapour in the remaining partA. (a)must be saturatedB. (b)must be unsaturatedC. (c)may be unsaturatedD. (d)there will be no vapour.

To what temperature should the hydrogen at `327^(@)C` be cooled at constant pressure, so that the root mean square velocity of its molecules become half of its previous value?A. `-123^(@)C`B. `23^(@)C`C. `-100^(@)C`D. `0^(@)`

A gas expands adiabatically at constant pressure such that its temperature `Tprop(1)/(sqrt(V))` , the value of `C_(P)//C_(V)` of gas isA. `1.30`B. `1.50`C. `1.67`D. `2.00`

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.

For adiabatic process of an ideal gas the value of `(dP)/P` is equals toA. `-gamma (dV)/V`B. `-gamma V/(dV)`C. `(dV)/V`D. `-gamma^(2) (dV)/V`

The coeffcient of performance of a refrigerator working between `10^(@)C` and `20^(@)C` isA. 28.3B. 29.3C. 2D. Cannot be calculated

On a cold winter night you switch on a room heater to keep your room warm. Does it mean that the total energy of the air inside the room increases after you switch on the heater?

The root mean square velocity of the molecules in a sample of helium is `5//7th` that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is `0^(@)C`, then the temperature of the helium sample is aboutA. `0^(@)C`B. 4 KC. `273^(@)C`D. `100^(@)C`

An engine has an efficiency of `1/6`. When the temperature of sink is reduced by `62^(@)C`, its efficiency is doubled. Temperature of the source isA. `127^(@)C`B. `37^(@)C`C. `62^(@)C`D. `99^(@)C`

In the above problem work done along path BC is:A. zeroB. (40 - 20 )= 20 JouleC. 40 JouleD. 60 Joule

A planet is at an average distance d from the sun and its average surface temeperature is T. Assume that the planet receives energy only from the sun and loses energy only through radiation from the surface. Neglect atmospheric effects. If `Tpropd^(-n)`, the value of n isA. 2B. 1C. `1//2`D. `1//4`

Find the external work done by the system in keal, when 20 kcal of heat is supplied to the system and the increase in the internal energy is 8400J (J=4200 J//kcal)?A. 16kcalB. 18kcalC. 20kcalD. 19 kcal

Calculate the kinetic energy of translation of the molecules of `20g` of `CO_(2)` at `27^(@)C`.

If three gas molecules have velocity `0.5`, `1` and `2 km//s` respectively, find the ratio of their root mean square speed and average speed.A. `V_(rms)=(V_(avg))/(2)`B. `V_(rms)=V_(avg)`C. `V(r.m.s)=1.134V_(av)`D. `V_(rms)=2V_(avg)`

Consider a gas at tmperature T occupying a volume V consisting of a mixture of two gases having. `N_a & N_b` atoms of masses `m_a & m_b` respsectively. Give an expression for the total pressure exerted by the gasA. `((N_a + N_b)kT)/(V)`B. `(1.5(N_a + N_b)kT)/(V)`C. `((N_a +N_b)RT)/(V)`D. none

Derive an expression for pressure exterted by an ideal gas?A. `((N_(a)+N_(b))kT)/(V)`B. `(1.5(N_(a)+N_(b))kT)/(V)`C. `((N_(a)+N_(b))RT)/(V)`D. none

A gas take part in two thermal processes in which it is heated from the same initial state to the same final temperature. The processes are shown on the `P-V` diagram by straight lines `1 rarr 3 ` and `1 rarr 2` A. `W_(13)gt W_(12)`B. `W_(12)gt W_(13)`C. `W_(13)= W_(12)`D. None of above

A gas take part in two thermal processes in which it is heated from the same initial state to the same final temperature. The processes are shown on the `P-V` diagram by straight lines `1 rarr 3 ` and `1 rarr 2` A. `Q_(13) lt Q_(12)`B. `Q_(13) gt Q_(12)`C. `Q_(13) = Q_(12)`D. None of above

Two temperature scales A and B are related by : `(A-42)/(110)=(B-72)/(220)` At which temperature two scales have the same reading ?A. `-42^(@)`B. `-72^(@)`C. `+12^(@)`D. `-40^(@)`

The work done on the system in an aidabatic compression depends onA. the increase in internal energy of the systemB. the decrease in internal energyC. the change in volume of the systemD. all the above

During adiabatic compression of a gas, its temperatureA. fallsB. raisesC. remains constantD. becomes zero

In the adiabatic compression the decrease in volume is associated withA. increase in temperature and decrease in pressureB. decrease in temperature and increase in pressureC. decreae in temperature and decrease in pressureD. increase in temperature and increae in pressure

For adiabatic processes `(gamma = (C_(p))/(C_(v)))`A. `P^(1-gamma)=` constantB. `P^(gamma)T^(1-gamma)=` constantC. `PT^(gamma)=` cosntantD. `P^(gamma)T=` constant

In an adiabatic process, `R = (2)/(3) C_(v)`. The pressure of the gas will be proportional to:A. `T^(5//3)`B. `T^(5//2)`C. `T^(5//4)`D. `T^(5//6)`

If an ideal gas is isothermally expanded its internal energy willA. increaseB. decreaseC. remains sameD. decrease or increase depending on nature of the gas

The temperature of the system decreases in the process ofA. free expansionB. isothermal expansionC. adiabatic expansionD. isothermal compression

In a adiabatic process is increased by `2//3%` if `C_(P)//C_(V) = 3//2`. Then the volume decreases by aboutA. `(4)/(9)%`B. `(2)/(3)%`C. `4%`D. `(9)/(4)%`

For an adiabatic process the relation between V and T is given byA. `TV^(gamma)=` constantB. `T^(gamma)V=`constantC. `TV^(1-gamma)=` constantD. `TV^(gamma-1)=`constant

The temperature of the system decreases in the process ofA. free expansionB. adiabatic expansionC. isothermal expansionD. isothermal compression

Heat engine rejects some heat to the sink. This heatA. converts into electrical energy.B. converts into light energy.C. converts into electromagnetic energyD. is unavailable in the universe.