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.
| 101. |
The Maxwell-Boltzmann distribution of molecular speeds in a sample of an ideal gas can be expressed as `f=(4)/(sqrt(pi))((m)/(2kT))^(3//2)v^(2)e^(-(mv^(2))/(2kT)).dv` Where f represent the fraction of total molecules that have speeds between v and v + dv.m, k and T are mass of each molecule, Boltzmann constant and temperature of the gas. (a) What will be value of `int_(v=0)^(v=oo)fdv ?` (b) It is given that `int_(0)^(oo) v^(3)e^(-av^(2))dv=(1)/(2a^(2))` Find the average speed of gas molecules at temperature T. |
|
Answer» Correct Answer - (a) 1 " " `(b) sqrt((8kT)/(pim))` |
|
| 102. |
Tidal volume is that volume of air which is inhaled (and exhaled) in one breath by a human during quiet breathing. For a normal man this volume is close to 0.6 litre. 35% of oxygen (in terms of molecular count) gets converted into carbon – di – oxide in the exhaled air. Nearly 21% of the air that we breath in is oxygen. Assume that a man is breathing when the air is at STP and the moisture content of inhaled and exhaled air is same. Estimate the difference in mass of an inhaled breath and exhaled breath. |
|
Answer» Correct Answer - 0.24 mg |
|
| 103. |
An ideal gas initially at 30 K undergoes an isobaric expansion at 2.50 k Pa. If the volume increases from `1.00 m^(2)` to `3.00 m^(3)` and 12.5 kJ is transferred to the gas by heat, what are (a) the change in its internal energy and (b) its final temperature ? |
|
Answer» The gas pressure is much less than one atmosphere. This could be managed by having the gas in a cylinder with a piston on the bottom end, supporting a constant load that hangs from the piston. The (large) cylineder is put into a warmer environment of make the gas expand. The first law of thermodynamics will tell us the change in internal energy. The ideal gas law will tells us the change in internal energy. The ideal gas law will tells us the final temperature. a. `Delta U - Q + W` where `W = - P Delta V` for a constant pressure process. So `Delta U = Q - P Delta V` `Delta U = 1.25 xx 10^(4) J - (2.50 xx 10^(3) N//m^(2))` `xx (3.00 m^(3) xx 10 m^(3)) = 7500 J` b. Since pressure and quantity of the gas are constant, we have from the equation of state `(V_(1))/(T_(1)) = (V_(2))/(T_(2))` and `T_(2) = ((V_(2))/(V_(1))) T_(1) = ((3.00 m^(3))/(1.00 m^(3))) (300 K) = 900 K` |
|
| 104. |
When an ideal gas is taken from state `a` to `b`, along a path `acb, 84 kJ` of heat flows into the gas and the gas does `32 kJ` of work. The following conclusions are drawn. Mark the one which is not correct. A. If the work done along the path `adb` is `10.5 kJ`, the heat will flow into the gas is `62.5 kJ`B. When the gas is returned from `b` to `a` along the curved path, the work done on the gas is `21 kJ`, and the system absorbs `73 kJ` of heatC. If `U_(a) = 0, U_(d) = 42 kJ`, and the work done along the path `adb` is `10.5 kJ` then the heat absorbed in the process `ad` is `52.4 kJ`.D. If `U _(a) = 0, U_(d) = 42 kJ`, heat absorbed in the precess `db` is `10 kJ`. |
|
Answer» Correct Answer - B b. For path `acb`: `Delta Q = Delta U + Delta W` `implies 84 = Delta U + 32 implies Delta U = 52 kJ` Hence `Delta U_(acb) = Delta U_(ab) = Delta U_(adb) = 52 J` For path `adb`: `Delta Q = Delta U + Delta W` `= 52 + 10.5 = 52.5 kJ` So option (a) is correct. For process `ba`, system will release the heat. So option (b) is wrong. For path `ad`: `Delta W_(adb) = Delta W_(ad) + Delta W_(db)` `implies 10.5 = Delta W_(ad) + 0` `implies Delta W_(ad) = 10.5 kJ` `Delta Q_(ad) = Delta U_(ad) + Delta W_(ad)` `= (42 - 0) + 10.5` `= 52.5 kJ` So option (c ) is correct. `Delta Q_(adb) = 52 + 10.5 = 62.5 kJ` `Delta Q_(db) = Delta Q_(adb) - Delta Q_(ad)` `= 62.5 52.5` `= 10 kJ` So option (d) is correct. Hence answer of this question is (b) |
|
| 105. |
Figure. Shows an ideal gas changing its state `A` to state `C` by two different path `ABC` and `AC`. a. Find the path along which the work done is the least. b. The internal energy of the gas at `A` is `10 J` and the amount of heat supplied to change its state to `C` through the path `AC` is `200 J`. Find the internal energy at `C`. c. The internal energy of the gas at state `B` is `20 J`. Find the amount of heat supplied to the gas to go from state `A` to state `B`. |
|
Answer» a. We know work done is given by the area below `PV` curve thus by observing two `PV` curves, `AC` and `ABC`, we can say that work done in path `AC` is less than that in path `ABC`, b. Work done in path `AC` by the gas is `W_(AC) =` area of `ACFEDA` = area of `ACF +` area of `AFED` `= (1)/(2) xx (15 - 5) xx (6 - 2) xx 5` `= 20 + 20 = 40 J` It is given that heat supplied in process `AC` is `Q_(AC) = 200 T` Thus change in intenal energy of gas in path `AC` is, from the first law of thermodynamics, given as `Q_(AC) = W_(AC) + Delta U_(AC)` or `Delta U_(AC) = Q_(AC) - W_(AC) = 200 - 40 = 160 J` As it is given that at state `A`, internal energy of gas is `10 J` thus at state `C`, internal energy is `Delta U_(AC) = U_(C ) - U_(A)` or `U_(C ) = Delta U_(AC) + U_(A) = 16 + 10 = 170 J` c. As in process `AB` no volume change takes place thus no work is done by or on the during path `AB`. Thus according to the first law of thermodynamics, we have `Q_(AB) = Delta U_(AB) + W_(AB)` `Q_(AB) = U_(B) + 0` or `Q_(AB) = 20 - 10 = 10 J` |
|
| 106. |
A gas is filled in the cylinder shown in fig. The two pistons are joined by a string. If the gas is heated, the right piston will A. move toward leftB. move toward rightC. remain stationaryD. none of these |
|
Answer» Correct Answer - B b. When temperature of gas increase, it expands. An the cross-section area of right is more, greater force will work on it (because `F = PA`). So piston will move towards rigth. |
|
| 107. |
At constant volume the molar heat capacity of oxyhydrogen gas (mechanical mixture of hydrogen and oxygen) is `n` times greater than that of water produced by the chemical combination of the gases. Find `n` |
|
Answer» Correct Answer - `5/6` |
|
| 108. |
There are two process `ABC` and `DEF`. In which of the process is the amount of work done by the gas is greater ? A. ABCB. DEFC. Equal in both processD. it cannot be predicted |
|
Answer» Correct Answer - B b. `W_(ABC) = (pi r^(2))/(2) = (pi (6)^(2))/(2) = 18 pi` `W_(DEF) = (pi xx 3^(2))/(2) + (pi xx 3^(2))/(4) + (15 - 12) xx 18` ` = 6.75 pi + 54` `W_(DEF) gt W_(ABC)` |
|
| 109. |
Air is filled at `60^(@)C` in a vessel of open mouth. The vessle is heated to a temperature `T` so that `1//4th` of air escapes. Assuming the volume of vessel remaining constant, the value of `T` isA. `80^(@)C`B. `440^(@)C`C. `333^(@)C`D. `171^(@)C` |
|
Answer» Correct Answer - D d. `M_(1) = M, T_(1) = 60 + 273 = 333 K` `M_(2) = M - (M)/(4) = (3 M)/(4)` (as `1//4th` part of air escapes) If pressure and volume of the gas remain constant, then `MT=` constant `:. (T_(2))/(T_(1)) = (M_(1))/(M_(2)) = ((M)/(3M//4)) = (4)/(3)` `implies T_(2) = (4)/(3) xx T_(1) = (4)/(3) xx 333 = 444 K = 171^(@)C` |
|
| 110. |
At a given temperature, the specific heat of a gas at constant pressure is always greater than its specific heat at constant volume.A. There is greater inter molecular atraction at constant pressureB. At constnat pressure molecular oscillations are more violentC. External work need to be done for allowing expansion of gas at constant pressure.D. Due to more reason other than those mentioned in the above |
| Answer» Correct Answer - C | |
| 111. |
Statement I: The specific heat of a gas in an adiabatic process is zwero but it is infinite in an isothermal process. Statement II: Specific heat of a gas is directly proportional to heat exchanged with the system and inversely proportional to change in termperature.A. Statement I: is true, Statement II is true and Statement II is the correct explanation for Statement I.B. Statement I: is true, Statement II is true and Statement II is NOT the correct explanation for Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
|
Answer» Correct Answer - C `C=(DeltatQ)/( M DeltaT)` In adiabatic process, `Delta Q=0. ` In isothermal process, `Delta T =0`. |
|
| 112. |
What is specific heat of gas in isothermal changes?A. infinityB. zeroC. negativeD. remains constant |
| Answer» Correct Answer - A | |
| 113. |
The gases have two principal specific heats but solids and liquied have only one specific heat. Why ?A. SolidB. GasC. LiquidD. Plasma |
| Answer» Correct Answer - B | |
| 114. |
Figure shows the adiabatic curve for `n` moles of an ideal gas, the bulk modulus for the gas corresponding to the point `P` will be A. `(5 nRT_(0))/(3 V_(0))`B. `nR (2 + (T_(0))/(V_(0)))`C. `nR (1 + (T_(0))/(V_(0)))`D. `(2 nRT_(0))/(V_(0))` |
|
Answer» Correct Answer - D d. For adiabatic process : Bulk modulus : `B = gamma P` for point `p : P = (nRT)/(V) = (nR 3 T_(0))/(3 V_(0)) = (nRT_(0))/(V_(0))` `implies B = (gamma nRT_(0))/(V_(0))` Now `TV^(gamma - 1) =` constant `implies (gamma - 1) TdV + VdT = 0` `implies (dV)/(dT) = (-V)/((gamma - 1) T)` for point `P rarr` `(-3 V_(0))/(3 T_(0)) = (- (3 V_(0)))/((gamma -1) (e T_(0))0` `implies gamma = 2` so from Eq. (i) `B = (2 nRT_(0))/(V_(0))` |
|
| 115. |
Find the number of atoms in molecule of a gas for which the ratio of specific heats at constant pressure and constant volume`lamda` becomes `(25)/(21)` times if the rotational degree of freedom of its molecules are frozen. Assume that the gas molecules originally had translational and rotational degree of freedom. |
|
Answer» Correct Answer - Two |
|
| 116. |
An ideal gas expands following a relation `(P^(2))/(rho)`= constant, where P = pressure and `rho` = density of the gas. The gas is initially at temperature T and density `rho` and finally its density becomes `(rho)/(3)`. (a) Find the final temperature of the gas. (b) Draw the P – T graph for the process. |
|
Answer» Correct Answer - (a) `sqrt(3) T` |
|
| 117. |
P-V curve of a diatomic gas is shown in the Fig. Find the total heat given to the gas in the process `A rarr B rarr C` |
|
Answer» From first law of thermodynamics `Delta Q_(ABC) = Delta U_(ABC) + Delta W_(ABC)` `Delta W_(ABC) = Delta W_(AB) + Delta W_(BC) = 0 + n RT_(B) 1n (V_(C ))/(V_(0)) = nRT_(B) 1n (2V_(0))/(V_(0))` `= nRT_(B) 1n2 = 2P_(0) V_(0) 1n2` `Delta U = nC_(v) Delta T = (5)/(2) (2P_(0) V_(0) - P_(0) V_(0))` `implies Delta Q_(ABC) = (5)/(2) P_(0) V_(0) + 2 P_(0) V_(0) 1n2` |
|
| 118. |
A gas is heated at a constant pressure. The fraction of heat supplied used of external work isA. `(1)/(gamma)`B. `(1 - (1)/(gamma))`C. `gamma - 1`D. `(1 - (1)/(gamma^(2)))` |
|
Answer» Correct Answer - B b. We know fraction of given energy that goes to increase the internal energy `= 1//gamma` So we can say the fraction of given energy that is supplied for external work `= 1 - (1//gamma)`. |
|
| 119. |
A sample of ideal gas `(gamma = 1.4)` is heated at constant pressure. If an amount `140 J` of heat is supplied to the gas, find (a) the changes in internal energy of the gas, (b) the work done by the gas. |
|
Answer» Suppose the simple contains `n` moles. Also suppose the volume changes from `V_(1)` to `V_(2)` and the temperature changes from `T_(1)` and `T_(2)`. The heat supplied is `Delta Q = Delta U + P Delta V = Delta U + n R Delta T = Delta U + (2 Delta U)/(f)` a. The change in internal energy is `Delta U = n (f)/(2) R (T_(2) - T_(1)) = (f)/(2) R n (T_(2) - T_(1))` `= (f)/(2 + f) Delta Q = (140 J)/(1.4) = 100 J` b. The work done by the gas is `Delta W = Delta Q - Delta U = 140 J - 140 J - 100 J = 49 J` |
|
| 120. |
One mole of an ideal gas has an interal energy given by `U=U_(0)+2PV` , where `P` is the pressure and `V` the volume of the gas. `U_(0)` is a constant. This gas undergoes the quasi`-` static cyclic process `ABCD` as shown in the `U-V` diagram. The molar heat capacity of the gas at constant pressure isA. `2R`B. `3R`C. `(5)/(2)R`D. `4R` |
|
Answer» Correct Answer - B `DeltaU=2 Delta(P V)=2 R DeltaT` `implies C_(V)=2R` `implies C_(P)=3R` |
|
| 121. |
One mole of an ideal gas has an interal energy given by `U=U_(0)+2PV` , where `P` is the pressure and `V` the volume of the gas. `U_(0)` is a constant. This gas undergoes the quasi`-` static cyclic process `ABCD` as shown in the `U-V` diagram. The gas must beA. monatomicB. diatomicC. a mixture of mono and diatomic gasesD. a mixture of di`-` and tri`-` atomic gases |
|
Answer» Correct Answer - C `C_(V)=2R`, which is between the `C_(V)` values of mono and diatomic gases. |
|
| 122. |
One mole of an ideal gas has an interal energy given by `U=U_(0)+2PV` , where `P` is the pressure and `V` the volume of the gas. `U_(0)` is a constant. This gas undergoes the quasi`-` static cyclic process `ABCD` as shown in the `U-V` diagram. The work done by the ideal gas in the process `AB` isA. zeroB. `(U_(1)-U_(0))/(2)`C. `(U_(0)-U_(1))/(2)`D. `(U_(0)-U_(1))/(2) log _(e)2` |
|
Answer» Correct Answer - D `W=RTlog_(e)(2V_(0))/(V_(0))=PVlog_(e)2=(U_(1)-U_(0))/(2) log_(e)2` |
|
| 123. |
Assertion : Mean free path of a gas molecule varies inversely as density of the gas. Reason : Mean free path varies inversely as pressure of the gas.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 The mean free path of a gas molecule is the average distance between two successive collisions. It is represented by `lambda` `lambda=(kT)/(sqrt(2)pisigma^(2)rho)` Hence, mean free path varies inversely as density of the gas. It can be easily proved that the mean free path varies directly as the temperature and inversely as the pressure of the gas. |
|
| 124. |
Six molecules speed 2 unit , 5 unit , 3unit, 6 unit, 3 unit , and 5unit , respectively . The rms speed isA. 4 unitB. 1.7 unitC. 4.2 unitD. 5 unit |
|
Answer» Correct Answer - C We have , `v_("rms")=sqrt((V_(1)^(2)+V_(2)^(2)+....+V_(n)^(2))/(n))` `=sqrt((4+25+9+36+9+25)/(6))` `sqrt((108)/(6))=sqrt(18)=3sqrt(2)=3xx1.414=4.242 ` unit |
|
| 125. |
Three copper blocks of masses `M_(1), M_(2) and M_(3) kg` respectively are brought into thermal contact till they reach equlibrium. Before contact, they were at `T_(1), T_(2), T)(3),(T_(1)gtT_(2)gtT_(3))`. Assuming there is no heat loss to the surroundings, the equilibrium temperature `T` is `(s is specific heat of copper)`A. `T=(T_(1)+T_(2)+T_(3))/(3)`B. `T=(M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3))/(M_(1)+M_(2)+M_(3))`C. `T=(M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3))/(3(M_(1)+M_(2)+M_(3)))`D. `T=(M_(1)T_(1)s+M_(2)T_(2)s+M_(3)T_(3)s)/(M_(1)+M_(2)+M_(3))` |
|
Answer» Correct Answer - B If the equilibrium temperature `TgtT_(1)andT_(2)` but less than `T_(3)` then as there is no heat loss to the surroundings therfore heat lost loss to the surroundings therfore heat lost by `M_(1)and M_(2)="heat gained by "M_(3)` `M_(1)s(T_(1)-T_(2))+M_(2)s(T_(2)-T)=M_(3)s(T-T_(3))` `M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3)=(M_(1)+M_(2)+M_(3))T` `rArr" "(M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3))/(M_(1)+M_(2)+M_(3))` |
|
| 126. |
An insulator container contains `4` moles of an ideal diatomic gas at temperature T. Heat Q is supplied to this gas, due to which `2` moles of the gas are dissociated into atoms but temperature of the gas remains constant. ThenA. `Q = 2RT`B. `Q = RT`C. `Q = 3RT`D. `Q = 4RT` |
|
Answer» Correct Answer - B `Q = Delta U =U_(f)-U_(i)` = [internal energy of 4 moles of a monoatomic gas + internal energy of 2 moles of a diatomic gas] - (internal energy of 4 moles of a diatomic gas] `=(4xx3/2RT+2xx5/2 RT)-(4xx5/2 RT) = RT` . |
|
| 127. |
Four moles of a perfect gas is heated to increaes its temperature by `2^(@)C` absorbs heat of 40cal at constant volume. If the same gas is heated at constant pressure find the amount of heat supplied. |
|
Answer» At constant volume `dQ=nC_(v)dT=dU=40` At constant pressure `dQ=dU+nRdT=40+(4xx2xx2)=56` cal |
|
| 128. |
A quantity of heat Q is supplied to a monoatomic ideal gas which expands at constant pressure. The fraction of heat that goes into work done by the gas `((W)/(Q))` is |
|
Answer» `C_(P)dT=C_(v)dT+dW, therefore dW=(C_(P)-C_(v))dT` Fraction of heat converted into work `(dW)/(dQ)=((C_(P)-C_(v))dT)/(C_(P)dT)=1-(C_(v))/(C_(P))=1-(1)/(gamma)` For monoatomic gas, `gamma=5//3` `therefore (dW)/(dQ)=1-(1)/(gamma)=1-(3)/(5)=(2)/(5)`, |
|
| 129. |
When an ideal diatomic gas is heated at constant pressure, the fraction of heat energy supplied which is used in doing work to maintain pressure constant isA. `5//7`B. `7//2`C. `2//7`D. `2//5` |
|
Answer» Correct Answer - C `dU=dQ-P(V_(2)-V_(1))` |
|
| 130. |
When an ideal diatomic gas is heated at constant pressure fraction of the heat energy supplied which increases the internal energy of the gas is |
|
Answer» Heat used in increasing the internal energy is `Q_(1)=C_(v)dT`, Heat absorbed at constant pressure to increase the temperature by `dT` is `Q_(2)=C_(P)dT` `therefore(Q_(1))/(Q_(2))=(C_(v))/(C_(P))=(1)/(C_(P)//C_(v))=(1)/(gamma)` for diatomic gas, `gamma=7//5, therefore (Q_(1))/(Q_(2))=(5)/(7)` |
|
| 131. |
When an ideal monoatomic gas is heated at constant pressure, fraction of heat energy supplied which increases the internal energy of gas , isA. `3//7`B. `5//7`C. `2//5`D. `3//5` |
|
Answer» Correct Answer - D For gas heated at constant pressure `Q = nC_(p)DeltaT=n 5/2 RDeltaT` Increase in internal energy `DeltaU=nC_(V)DeltaT=n.3/2RDeltaT` `:.` fractional of heat energy supplied = `(DeltaU)/(Q)` `=((3//2)nRT)/((5//2)nRT) = 3/5` |
|
| 132. |
A sphere, a cube and a thin circular plate all made of the same material and having the same mass are initially heated to a temperature of `300^(@)C`. Which one of these cools faster ?A. SphereB. CubeC. PlateD. None of these |
|
Answer» Correct Answer - C The rate of cooling is directly proportional to surface area. |
|
| 133. |
A sample of an ideal gas is taken through the cyclic process `abca` . It absorbs `50 J` of heat during the part `ab`, no heat during `bc` and rejects `70 J` of heat during `ca`. `40 J` of work is done on the gas during the part `bc`.(a) Find the internal energy of the gas at `b` and `c` if it is `1500 J` at `a`.(b) Calculate the work done by the gas during the part `ca`. |
|
Answer» a. In the part `ab` the volume remains constant. Thus, the work done by the gas is zero. The heat absorbed by the gas is `50 J` The increase in internal energy from a to b is `Delta U = Delta Q = 50 J` As the internal energy is `1500 J` at `a`, it will be `1500 J` at `b`. In the part `bc`, the work done by the gas is `Delta W = - 40 J` and no heat is given to the system. The increase in internal energy from b to c is `Delta U - Delta W = 40 J` As the internal energy is `1550 J` at `b`, it will be `1590 J` at c. b. The change in internal energy from c to a is `Delta U = 1500 J - 1590 J = - 90 J` The heat given to the system is `Delta Q = - 70 J`. Using `Delta Q = Delta U + Delta W`, `Delta W ca = Delta Q - Delta U = - 70 J + 90 J = 20 J` |
|
| 134. |
Consider the cyclic process ABCA, shown in figure, performed on a sample of `2.0 mol` of an ideal gas. A total of `1200 J` of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC. A. `2580 J`B. `3625 J`C. `4520 J`D. `1550 J` |
|
Answer» Correct Answer - C `W_(AB) = nRT Delta T = 400R` `W_(CA) = 0` For the cyclic process `ABCA` `W_(AB) +W_(BC) +W_(CA) = Delta Q = -1200J` `implies W_(BC) = -1200-400R =-1200-3320 = -4520J` |
|
| 135. |
If the ratio of specific heat of a gas of constant pressure to that at constant volume is `gamma`, the change in internal energy of the mass of gas, when the volume changes from `V` to `2V` at constant pressure `p` is |
|
Answer» `dU=nC_(v)dT=n(R)/(gamma-1)dT=n(PdV)/(gamma-1)` `=n(p(2V-V))/(gamma-1)=(PV)/(gamma-1)` |
|
| 136. |
Three moles of an ideal monoatomic gas per form a cyclic as shown in the Fig. the gas temperature in different states are `T_(1) = 400 K, T_(2) = 800 K, T_(3) = 2400 K and T_(4) = 1200 K`. The work done by the gas during the cyclic is A. `10 kJ`B. `20 kJ`C. `5 kJ`D. `8.3 kJ` |
|
Answer» Correct Answer - B In the curves `1-2` and `3-4` we find that the pressure is directly proportional to temperature. So the volume remains unchanged. i.e., gas does not work . The work done during the isobaric processes `2-3` and `1-4` are follows: `W_(2-3) = P_(2)(V_3-V_2)` `W_(1-4) = P_(1)(V_1-V_4)` Total work done `=P_(2)(V_3-V_2)+P_(1) (V_1-V_2)` `:. W_T = 3RT_3-3RT_2+3RT_1-3RT_4` Three moles has been given, so `PV = nRT = 3RT` `:. W_T = 3RT_(3) -3RT_(2)+3RT_1-3RT_4` `= 3R[400+2400-800-1200]` `=3R xx 800 = 20xx1^(3) J = 20kJ` . |
|
| 137. |
Consider the cyclic process ABCA, shown in figure, performed on a sample of `2.0 mol` of an ideal gas. A total of `1200 J` of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC. |
|
Answer» In a cyclic process `Delta U = 0` From the first law of thermodynamics, for the cyclic process `Q = Delta U + W` `W = Q - Delta U = - 1200 - 0` `= - 1200 J` From `C` to `A`, `Delta v = 0` `:. W_(CA) = 0` For the whole cycle `W_(AB) + W_(BC) + W_(CA) = W` `= 0 1200 J` As `W_(CA) = 0` `:. W_(AB) + W_(BC) = - 1200 J` Work done form `A` to `B` : In the process `V poro T`, so pressure remains constant we know that `PV = n RT` or `P Delta V = nR Delta T` `:. W_(AB) = P Delta V = nR Delta T` `= 2 xx 8.31 xx (500 - 300) = 3324 J` Subsituting this value in Eq. (i) we get `3324 + W_(BC) = - 1200` `W_(BC) = - 4524 J` |
|
| 138. |
Three moles of an ideal monoatomic gas undergoes a cyclic process as shown in the figure. The temperature of the gas in different states marked as 1,2,3 and 4 are 400K, by the gas during the process 1-2-3-4-1 is (universal gas constant is R) |
|
Answer» Process `1rarr2` and `3rarr4` are polytrophic and process `2rarr3` and `4rarr1` are isobaric. From the graph `PproprArr(P)/(V)=KrArrPV^(-1)=KrArrx=-1` Work done `W=W_(1rarr2)+W_(2rarr3)+W_(3rarr4)+W_(4rarr1)` `=(nR)/(x-1)(T_(1)-T_(2))+P_(2)(V_(3)-V_(2))+(nR)/(x-1)[T_(3)-T_(4)]+P_(1)[V_(1)-V_(4)]` `(nR)/(x-1)[T_(1)-T_(2)]+nR(T_(3)-T_(2))+(nR)/(x-1)[T_(3)-T_(4)]+nR[T_(1)-T_(4)]` `=(nR)/(x-1)[T_(1)-T_(2)+T_(3)-T_(4)]+nR(T_(3)-T_(3)+T_(1)-T_(4))``=(nR)/(x-1)[T_(1)-T_(2)+T_(3)-T_(4)]+nR(T_(3)-T_(3)+T_(1)-T_(4))` `=(3R)/(-1-1)[400-700+2500-100]` `+3R(2500-700+400-1100)` `=(3R)/(-1-1)[1100]+3R(1100)` `=3R(1100)(1-(1)/(2))=1650R` |
|
| 139. |
Figure, shows the variation of potential energy `(U)` of 2 mol of Argon gas with its density in a cyclic process `ABCA`. The gas was initiallly in the state `A` whose pressure and temperature are `P_(A)=2 atm, T_(A)=300K`, respectively. It is also stated the path `AB` is a rectangular hyperbola and the internal energy of the gas at state `C` is `3000r`. Based on the above information answer the following question `:` The heat supplied to the gas in the process `AB` isA. `700R`B. `3500R`C. `4400R`D. `1600R` |
|
Answer» Correct Answer - B Given `U_(C)=3000R=Nc_(v)T_(C)` `implies 3000 r = 2 XX (3)/( 2) R xx T_(C)` `implies T_(C)=1000K` Also `U_(B)=U_(C)` and `U_(A)= 2 xx(3)/(2) Rxx 300=900 R` `W_(AB)=PDeltaV=n RDeltaT` `=2 xx R(1000-300)` `=1400R` `U_(AB)=U_(B)-U_(A)=3000R-900R` `=2100R` `:. DeltaQ=DeltaW+DeltaU` `=1400R+2100R=3500R` |
|
| 140. |
Figure, shows the variation of potential energy `(U)` of 2 mol of Argon gas with its density in a cyclic process `ABCA`. The gas was initiallly in the state `A` whose pressure and temperature are `P_(A)=2 atm, T_(A)=300K`, respectively. It is also stated the path `AB` is a rectangular hyperbola and the internal energy of the gas at state `C` is `3000r`. Based on the above information answer the following question `:` A. The process `AB` is isobaric, `BC` is adiabatic and `CA` is isochoric.B. The pocess is `AB` is adiabatic, `BC` is isothermal and `CA` is isochoric.C. The process `AB` is isochoric, `BC` is isothermal and `CA` is isobaricD. The process `AB` is isochoric, `BC` is isothermal and `CA` is isochoric |
|
Answer» Correct Answer - D Since `AB` is rectangular hyperbola, therefore, `:. U rho=` Constant `implies n C_(v)T(PM)/(RT)=C` `implies P=` Constant `:. AB rarr` isobaric ` BC rarr` isothermal as `U` is constant `CA rarr` isochoric as `rho` is constant |
|
| 141. |
The volume `(V)` of a manatomic gas varies with its temperature `(T)` , as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state `A` to `B` , is A. `2/7`B. `2/5`C. `1/3`D. `2/3` |
|
Answer» Correct Answer - B `V prop T` Isobaric process `W = PDeltaV = muR DeltaT` `Delta Q = mu C_(P)Delta T` For monoatomic gas `f = 3` `C_(P)=(f/2 R+R) = 5/3R` So, `(W)/(Delta Q) = (muRDeltaT)/(mu C_(P)Delta T) = 2/5`. |
|
| 142. |
`3` moles of an ideal mono atomic gas performs a cycle as shown in fig. `If` gas temperature `T_(A)=400 K` `T_(B)=800K,T_(C)=2400K`, and `T_(D)=1200K`. Then total work done by gas is |
|
Answer» Processes A to B and C to D are parts os straightline graphs of forms y=mx and `P=(nR)/(V)T (n=3) i.e. PaT` So, volume remains constant for the graphs AB and CD. So, no work is done during processes for A to B and C to D. `W_(AB)=W_(CD)=0` and `W_(BC)=P_(2)(V_(C)-V_(B))` `=nR(T_(C)-T_(B))=3R(2400-800)=4800R` `W_(DA)=P_(1)(V_(A)-V_(D))=nR(T_(A)T_(D))` `=3R(400-1200)=-2400R` Work done in the complete cycle `W=W_(AB)+W_(BC)+W_(CD)+W_(DA)` `=0+4800R+0+(-2400)R` `=2400R=19953.6 Japprox20KJ` |
|
| 143. |
`P - V` diagram of a cyclic process `ABCA` is as shown in Fig. Choose the correct alternative A. `Delta Q_(A) rarr._(B)` is negativeB. `Delta U_(b) rarr._(c)` is negativeC. `Delta U_(c) rarr._(A)` is negativeD. `Delta W_(CAB)` is negative |
|
Answer» Correct Answer - A::B::D During process `A` and `B`, pressure and volume both are decreasing. Therefore , temperature and hence internal energy of the gas will decrease `(T prop PV)` or `Delta V_(A) rarr _(B)=` negative. Further, `Delta W_(A) rarr _(B)` is negative. In processs `B` to `C`, pressure of the gas is constant while volume is increasing . Hence, temperature should increase or `DeltaU_(B) rarr _(C) =` postive. During `C` to `A` volume is constant while pressure is increasing. Therefore, temperature and hence internal energy of the gas should increase or `Delta U _(C) rarr _(A)=` positive. During process `CAB`, volume of the gas is decreasing. Hence, work done by the gas is negative. |
|
| 144. |
A cyclic process `ABCA` is shown in the `V-T` diagram process on the `P-V` A. B. C. D. |
|
Answer» Correct Answer - C c. Form the given `V- T` diagrams, we can seed that in process `AB, V prop T`. Therefore pressure is constant (as quantity of the gas remains same). In process `BC, V` = constant and in process `CA` ` T = ` constant. Therefore these processes are correctly represented on `P-V` diagram by graph `(c )`, |
|
| 145. |
A bubble of 8 moles of helium is submerged at a certain depth in water. The temperature of water increases by `30^(@)C` How much heat is added approximately to helium during expansion ?A. 4000 JB. 3000 JC. 3500 JD. 4500 J |
|
Answer» Correct Answer - B Heat added to helium during expansion , `H=nC_(v)DeltaT=8xx(3)/(2)Rxx30(C_(V)" for monoatomic gas" =(3)/(2)R)` =360R 360xx8.31 ` (R=8.31J"mol"^(-1)-K^(-1))` `H~~3000J` |
|
| 146. |
`3` moles of an ideal mono atomic gas performs a cycle as shown in fig. `If` gas temperature `T_(A)=400 K` `T_(B)=800K,T_(C)=2400K`, and `T_(D)=1200K`. Then total work done by gas is |
|
Answer» `1rarr2` and `3rarr4` are isochoric processes. Therefore, work done is zero. `therefore W_("total")=W_(23)+W_(41)=P_(2)(V_(3)-V_(2))+P_(4)(V_(1)-V_(4))` `=nR(T_(3)-T_(2))+nR(T_(1)-T_(4))` `=nR(T_(3)-T_(2)+T_(1)-T_(4))=800nR=2400R` |
|
| 147. |
Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of heat absorbed and the work done by the gas in process `1-2` is |
|
Answer» V-T graph is a straight line passing through origin. Hence, `VpropT` or P=cosntant `(dQ)/(dW)=(nC_(P)DeltaT)/(nC_(P)DeltaT-n_C(v)DeltaT)=(nC_(P)DeltaT)/(nRDeltaT)=(5R)/(2R)=(5)/(2)` |
|
| 148. |
A cyclic process `ABCD` is shown is shown in the following `P-V` diagram. Which of the following curves represent the same process ? A. B. C. D. |
|
Answer» Correct Answer - A a. `AB` is isobaric process, `BC` is isothermal process, `CD` is isochoric process and `DA` is isothermal process, These processes are correctly represented by graph (a). |
|
| 149. |
Two moles of helium gas are taken along the path `ABCD` (as shown in Fig.) The work done by the gas is A. `2000 R(1+"in"(4)/(3))`B. `500 R(3 + "in" 4)`C. `500 R(2 + "In"(16)/(9))`D. `1000 R(1 + "In"(16)/(9))` |
|
Answer» Correct Answer - C `A rarr B` is an isobaric process `so, DeltaW_(AB) = nRDeltaT = 2 xx R xx (750-250)=1000R` `B rarr C` is an isochoric process `:. DeltaW_(BC)=0` and `C rarr D` is an isothermal process, `DeltaW_(CD)=nRT 1n ((V_f)/(V_i))` `=2xxRxx1000 1n(20/15)=2000 R 1n(4//3)` Total work done `Delta W = sigmaDeltaW_(AB)`. |
|
| 150. |
`P - V` diagram of an ideal gas is as shown in figure. Work done by the gas in process `ABCD` is |
|
Answer» `W_(AB)=-P_(0)V_(0)`, `W_(BC)=0` and `W_(CD)=AP_(0)V_(0)` `W_(ABCD)=-P_(0)V_(0)+0+4P_(0)V_(0)=3P_(0)V_(0)` |
|