4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 5 InterviewSolution

201.	A closed and isolated cylinder contains ideal gas. An adiabatic separator of mass m, cross sectional area `A` divides the cylinder into two equal parts each with volume `V_(0)` and pressure `P_(0)` in equilibrium Assume the sepatator to move without friction. If the piston is slightly displaced by x, the net force acting on the piston isA. `(P_(0)gammaA^(2)x)/(V_(0))`B. `(2P_(0)gammaA^(2)x)/(V_(0))`C. `(3P_(0)gammaA^(2)x)/(V_(0))`D. `(P_(0)gammaA^(2)x)/((gamma-1)V_(0))`
Answer» Correct Answer - B `DeltaQ=0` for system `P_(0)V_(0)^(gamma)=P_(1)V_(1)^(gamma)=P_(2)V_(2)^(gamma)` `impliesP_(1)=P_(0)[1-(gammaAx)/(V_(0))]impliesP_(2)=P_(0)[1+(gammaAx)/(V_(0))]` ` implies F = (2P_(0)gammaA^(2)x)/(V_(0)`

Discussion

202.	A gaseous mixture enclosed in a vessel consists of one gram mole of a gas A with `gamma=(5/3)` and some amount of gas B with `gamma=7/5` at a temperature T. The gases A and B do not react with each other and are assumed to be ideal. Find the number of gram moles of the gas B if `gamma` for the gaseous mixture is `(19/13)`.
Answer» Correct Answer - B As for ideal gas `C_(P)-C_(V)=R andgamma=(C_(P)//C_(V))` , `gamma-1=(R)/(C_(V)) orC_(V)=(R)/((gamma-1))` `(C_(V))_(1)=(R)/((5//3)-1)=(3)/(2)R` `(C_(V))_(2)=(R)/((7//5)-1)=(5R)/(2)` and `(C_(V))_(mix)=(R)/((19//13)-1)=(13)/(6)R` Now from the conservation of energy,i.e., `DeltaU=DeltaU_(1)+DeltaU_(2)` , we get `(mu_(1)+mu_(2))(C_(V))_(mix)DeltaT=[mu_(1)(C_(V))_(1)+mu_(2)(C_(V))_(2)]DeltaT` `(V_(V))_(mix)=(mu_(1)(C_(V))_(1)+mu_(2)(C_(V))_(2))/(mu_(1)+mu_(2))` `(13)/(6)R=(1xx(3)/(2)R+mu_(2)xx(5)/(2)R)/(1+mu_(2))=((3+5mu_(2))R)/(2(1+mu_(2)))` or `13+13mu_(2)=9+15mu_(2)` ,i.e.,`mu_(2),=2g "mole"`

Discussion

203.	If a gas of volume 10 L is expanded to quadruple its volume at 4 atm pressure, the external work done isA. 3 kJB. 6 kJC. 12 kJD. 18 kJ
Answer» Correct Answer - C

Discussion

204.	5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be `T_1,` the work done in the process isA. `(9)/(8)RT_(1)`B. `(3)/(2)RT_(1)`C. `(15)/(8)RT_(1)`D. `(9)/(2)RT_(1)`
Answer» Correct Answer - A `W=(nR(T_(1)-T_(2)))/(r-1) T_(1) (5.6)^(2//3)=T_(2)(0.7)^(2//3)` `impliesT_(2)=T_(1)(8)^(2//3)` `=T_(1)xx4 W =(nRxx3T_(1))/(2//3)=(9)/(2) nRT_(1)` But `n =(1)/(4) implies W = (9)/(8)RT_(1)`.

Discussion

205.	A sample of ideal gas `(gamma=1.4)` is heated at constant pressure. If an amount of 100 J heat is supplied to the gas, the work done by the gas isA. 28.57 JB. 56.54 JC. 38.92 JD. 65.38 J
Answer» Correct Answer - A `dQ=n C_(p)dT` `dU=nC_(V)dT` `dW=dQ-dU=n(C_(p)-C_(V))dT` `(dW)/(dQ)=(nC+_(p))-(C_(V)dT)/(nC_(p)dT)=(C_(p))/(C_(p))-(C_(V))/(C_(P))` `(dW)/(100)=1-(C_(V))/(C_(p))=1-(1)(1.4)=(0.4)/(1.4)=(4)/(14)=(2)/(7)` `dW=100xx(2)(7)=28.57 J`

Discussion

206.	The amount of heat supplied to `4xx10^(-2)` kg of nitrogen at room temperature to rise its temperature by `50^(@)C` at constant pressure is (Molecualr mass of nitrogen is 28 and `R= 8.3 J mol^(-1)K^(-1))`A. 2.08 KJB. 3.08 KJC. 4.08 KJD. 5.08 KJ
Answer» Correct Answer - A Given , `m=4xx10^(-2)kg =40 g. DeltaT=50^(c )` Number of moles ,`n=(m)/(M)=(40)/(28)=1.43` As nitrogen is a diatomic gas, molar specific heat at contstant pressure is `C_(p)=(7)/(2)R=(7)/(2)xx8.3 J mol^(-1) K^(-1)=29.05 J mol^(-1) K^(-1)` As `DeltaQ =nC_(p)DeltaT` `DeltaQ=1.43xx29.05xx50=2.08xx10^(3)` J =2 2.08KJ

Discussion

207.	`1672cal` of heat is given to one mole of oxygen at `0^(@)C` keeping the volume constant. Raise in temperature is `(c_(p)=0cal//gm^(0)k and R=2cal//"mole"//K)`A. `33.6^(@)C`B. `36.3^(@)C`C. `63.3^(@)C`D. `334.4^(@)C`
Answer» Correct Answer - D `DeltaQ_(V)=nC_(V)DeltaT`

Discussion

208.	`0.5` mole of diatomic gas at `27^(@)C` is heated at constant pressure so that its tripled. If `=8.3J "mole"^(-1)k^(-1)` then work done isA. `4980J`B. `2490J`C. `630J`D. `1345J`
Answer» Correct Answer - B `dQ_(V)=PdV=nRDeltaT`

Discussion

209.	Two metal balls having masses `50g` and `100g` collides with a target with same velocity. Then the ratio of their rise in temperature isA. `1:2`B. `4:1`C. `2:1`D. `1:1`
Answer» Correct Answer - D `(1)/(2)mv^(2)=JmSDelta thetaimpliesDelta0alphav^(2)`

Discussion

210.	A calorimeter of mass m contains an equal mass of water in it. The temperature of the water and clorimeter is `t_(2).A` block of ice of mass m and temperature `t_(3)lt0^(@)C` is gently dropped into the calorimeter. Let `C_(1),C_(2) and C_(3)` be the specific heats of calorimeter, water and ice respectively and `L` be the latent heat of ice. The whole mixture in the calorimeter becomes water ifA. `(C_(1)+C_(2))t_(2)-C_(3)t_(3)+Lgt0`B. `(C_(1)+C_(2))t_(2)+C_(3)t_(3)+Lgt0`C. `(C_(1)+C_(2))t_(2)-C_(3)t_(3)-Lgt0`D. `(C_(1)+C_(2))t_(2)+C_(3)t_(3)-Lgt0`
Answer» Correct Answer - D Heat lost `=mC_(1)(t_(2)-0)+mC_(2)(t_(2)-0)` Heat gained `=mC_(3)(0-t_(3))+mL` `:. C_(1)t_(2)+C_(2)t_(2)gtC_(3)t_(2)gtC_(3)t_(3)+L` `:. C_(1)t_(1)+C_(2)t_(2)+C_(3)t_(3)-L gt 0`

Discussion

211.	A calorimeter of mass m contains an equal mass of water in it. The temperature of the water and clorimeter is `t_(2).A` block of ice of mass m and temperature `t_(3)lt0^(@)C` is gently dropped into the calorimeter. Let `C_(1),C_(2) and C_(3)` be the specific heats of calorimeter, water and ice respectively and `L` be the latent heat of ice. Water equivalent of calorimeter isA. `mC_(1)`B. `(mC_(1))/(C_(2))`C. `(mC_(2))/(C_(1))`D. None of these
Answer» Correct Answer - B `mC_(1)=MC_(2) :. M=(mC_(1))/(C_(2))`

Discussion

212.	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))/(3(M_(1)+M_(2)+M_(3))`C. `T= (M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(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 tepm. `T gt T_(1)` and `T_(2)` but less than `T_(3)`, then as there is no heat loss to the surroundings, therfore, heat lost by `M_(1)` and `M_(2)` = heat gained by `M_(3)` `M_(1)s(T_(1)-T)+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` or `T=(M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3))/(M_(1)+M_(2)+M_(3))`

Discussion

213.	Three copper blocks of masses `M_(1), M_(2) and M_(3)` kg respectively are brought into thermal contact till they each equilibrium. 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 Let the equilibrium temperature of the system is T. Let us assume that `T_(1),T_(2) ltTltT_(3)`. ltBrgt According to question, there is no net loss to the surroundings. Heat lost by `M_(3)`= Heat gained by ` M_(1)` + Heat gained by `M_(2)` ltBrgt `rArr" "M_(3)s(T_(3)-T)=M_(1)s(T-T_(1))+M_(2)s(T-T_(2))` `" "` (where, s is specific heat of the copper material) `rArr" "T[M_(1)+M_(2)+M_(3)]=M_(3)T_(3)+M_(1)T_(1)+M_(2)T_(2)` `rArr" "T=(M_(1)T_(1)+M_(2)T_(2)+M_(3)T_(3))/(M_(1)+M_(2)+M_(3))`

Discussion

214.	Which of the processes described below are irreversible ?A. The increase in temperature of an iron rod by hammering itB. A gas in a small container at a temperature `T_(1)` is brought in contact with a big reservoir at a higher temperature `T_(2)` which increases the temperature of the gasC. A quasi-static isothermal expansion of an ideal gas in cylinder fitted with a frictionless pistonD. An ideal gas is enclosed in a piston cylinder arrangement with adiabatic walls. A weight w is added to the piston, resulting in compression of gas
Answer» Correct Answer - A::B::D (a) When the rod is hammered, the external work is done on the rod which increases its temperature. The process cannot be retracted itself. (b) In this process energy in the form of heat is transferred to the gas in the small container by big reservoir at temperature `T_(2)`. (d) As the weight is added to the cylinder arrangement in the form of external pressure hence, it cannot be reversed back itself.

Discussion

215.	A gas is compressed adiabatically till its temperature is doubled. The ratio of its final volume to initial volume will beA. `1/2`B. more then `1/2`C. less then `1/2`D. Between 1 and 2
Answer» Correct Answer - C

Discussion

216.	A tyre filled with air,`(27^(@)C`,o and 2 atm ) bursts, then what is temperature of air `(lambda=1.5)`A. `-33^(@)C`B. `0^(@)C`C. `27^(@)C`D. `240^(@)C`
Answer» Correct Answer - A

Discussion

217.	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 = 2RTB. Q = RTC. Q = 3RTD. Q = 4RT
Answer» Correct Answer - B

Discussion

218.	In a thermodynamic process on an ideal diatomic gas, work done by the gas is `eta` times. The heat supplied `(eta lt 1)`. The molar heat capacity of the gas for the process isA. `(5R)/(2)`B. `5eta(R)/(2)`C. `(5R)/(2(1-eta))`D. `(5R(1-eta))/(2)`
Answer» Correct Answer - C

Discussion

219.	When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied, which increases the internal energy of the gas, isA. `2/5`B. `3/5`C. `3/7`D. `5/7`
Answer» Correct Answer - D

Discussion

220.	One mole of an ideal gas at `27^@C` expanded isothermally from an initial volume of `1 ` litre to `10` litre. The `DeltaU` for this process is : `(R=2 cal K^(-1) mol^(-1))`A. zeroB. `9` litre-atmC. `1281.1 cal`D. `164.7 cal`
Answer» Correct Answer - A For isothermal process `DeltaI=0`

Discussion

221.	Enerrgy can transfter from syetm to surroundings as work ifA. There is thermal equilibrium between system and surrounding.B. There is mechanical equilibrium between system and surrounding.C. If pressute of system gt atmospheric pressure.D. None of these.
Answer» `w = P_(ex)(DeltaV)`

Discussion

222.	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 A. 3B. `5/2`C. `5/3`D. `7/2`
Answer» Correct Answer - B

Discussion

223.	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 A. `3`B. `5//2`C. `5//3`D. `7//2`
Answer» Correct Answer - B `V-T` graph is a straight line passing through origin.Hence, `VpropT` or `P` =constant `:. Dq=nC_(p)dT and dU=nC_(v)dT` `:. dW=dQ-dU=n(C_(p)-C_(v))dT` `:. (dQ)/(dW)=(nC_(p)dT)/(n(C_(p)-C_(v))dT)=(C_(p))/(C_(p)-C_(v))=(1)/(-1(C_(v))/(C_(p)))` `(C_(v))/(C_(P))=(3)/(5)` `:. (dQ)/(dW)=(1)/(1-3//5)=(5)/(2)`.

Discussion

224.	The door of a running refrigerator inside a room is left open. The correct statement out of the following ones isA. The room will be cooled slightltB. The room will be warmed up graduallyC. The room will be cooled to the temperature inside the refrigeratorD. The temperature of the room will remain uneffected.
Answer» Correct Answer - B Cofficient of performance of refrigerator is `(Q_(2))/W= (T_(2))/(T_(1)-T_(2))` When the door of a running refrigerator inside a room is left open, `T_(2)rarrT_(1)`. Coefficient of performance increases. `Q_(2)` increases. Therefore, heat energy given to the room increases. Hence the room will be warmed up gradually.

Discussion

225.	With what pressure must a given volume of oxygen, originally at 300K and 1 bar pressure be adiabatically and irreversibly compored in order to raise its temperture to 600K `(gamma=7//5)`A. `4.5 "bar"`B. `9"bar"`C. `2.25"bar"`D. `6.75"bar"`
Answer» Correct Answer - A

Discussion

226.	2 mole of ideal gas expands isothermically and reversibally from 1 L to 10 L at 300 K. then `DeltaH` is :A. 4.98 kJB. 11.47 kJC. `-11.47 kJ`D. zero
Answer» Correct Answer - D

Discussion

227.	`1dm^(3)` of an ideal gas at a pressure of 10bar expands reversibly and isothermally to final volume of 10 litre .(In 10=2.3). Heat absorbed in the process will be:A. `1.15kJ`B. `4.6J`C. `2.3kJ`D. `9.2kJ`
Answer» Correct Answer - C

Discussion

228.	For a chemical reaction the values of ∆H and ∆S at 300 K are – 10 kJ mol-1 and -20 J mol-1 respectively. What is the value of AG of the reaction? Calculate the ∆G of a reaction at,600K assuming ∆H and AS values are constant. Predict the nature of the reaction.
Answer» Given, ∆H =- 10 kJ mol = -10000 J mol^-1 ∆S = -20 JK^-1 mol^-1 T = 300 K ∆G = ? ∆G = ∆H – T∆S ∆G = – 10 kJ mol^-1 -300 K x (-20 x 10^-3) kJ K^-1 mol^-1 ∆G = (-10 + 6) kJ mol^-1 ∆G =-4 kJ mol^-1 At 600 K, ∆G = – 10 kJ mol^-1 -600 K x (-20 x 10^-3) kJ K^-1 mol^-1 ∆G =(-10+12) kJ mol^-1 ∆G = +2 kJ mol^-1 The value of ∆G is negative at 300K and the reaction is spontaneous, but at 600K the value ∆G becomes positive and the reaction is non-spontaneous.

228.

For a chemical reaction the values of ∆H and ∆S at 300 K are – 10 kJ mol-1 and -20 J mol-1 respectively. What is the value of AG of the reaction? Calculate the ∆G of a reaction at,600K assuming ∆H and AS values are constant. Predict the nature of the reaction.

Answer»

Given,

∆H =- 10 kJ mol = -10000 J mol^-1

∆S = -20 JK^-1 mol^-1

T = 300 K

∆G = ?

∆G = ∆H – T∆S

∆G = – 10 kJ mol^-1 -300 K x (-20 x 10^-3) kJ K^-1 mol^-1

∆G = (-10 + 6) kJ mol^-1

∆G =-4 kJ mol^-1

At 600 K,

∆G = – 10 kJ mol^-1 -600 K x (-20 x 10^-3) kJ K^-1 mol^-1

∆G =(-10+12) kJ mol^-1

∆G = +2 kJ mol^-1

The value of ∆G is negative at 300K and the reaction is spontaneous, but at 600K the value ∆G becomes positive and the reaction is non-spontaneous.

Discussion

229.	Calculate the work done (in joules) when 0.2 mole of an ideal gas at 300 K expands isothermally and reversibly from initial volume of 2.5 litres to the final volume of 25 litres.A. 996B. 1148C. 11.48D. 897
Answer» Correct Answer - b ` w = - 2.303nRT log. (V_(2))/(V_(1))= - 2.303 xx 0.2 xx 8.314 xx 300 log. ( 25)/( 2.5) = - 1148 J`

Discussion

230.	Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?
Answer» We have q = - w : pext (10 - 2) = 0 × 8 = 0 No. work done; No heat, is absorbed.

230.

Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?

Answer»

We have q = - w : pext (10 - 2) = 0 × 8 = 0

No. work done; No heat, is absorbed.

Discussion

231.	Two litres of an ideal gas at a pressure of 10 atm expands isothermally at `25^(@)C` into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?
Answer» We have `q=-w=p_(ex)(10-2)=0(8)=0` No work is done, no heat is absorbed.

Discussion

232.	Two litres of an ideal gas at a pressure of 10 atm expands isothermally at `25^(@)C` into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?A. 10 J, 10 JB. 8 J, 10 JC. 18 J, 0 JD. 0 J, 0 J
Answer» Correct Answer - D `q=-w=P_(ex)(10-2)=0(8)=0` For isothermal expansion in vacuum, `P_(ex)=0`

Discussion

233.	What is the entropy change when 1 mole oxygen gas expands isothermally and reversibly from an initial volume of 10 L to 100 L at 300 K ?A. `"19.14 J K"^(-1)`B. `"109.12 J K"^(-1)`C. `"29.12 J K"^(-1)`D. `"10 J K"^(-1)`
Answer» Correct Answer - A `DeltaS=2.303nR log .(V_(2))/(V_(1))` `=2.303xx1xx8.314log.(100)/(10)="19.14 J K"^(-1)`

Discussion

234.	Two litres of an ideal gas at a pressure of 10 atm expands Isothermally into vacuum until its total volume is 10 litres. How muach heat is absorbed and howmuch work is done in the expansion ? What would be the heat absorbed andwork done (i) if the same expansion takes place against a constant external pressure of 1 atm ? (ii) if the same expansion takes place to a final volume of 10 litres conducted reversibly?
Answer» For isothermal expansion of an ideal gas,` Delta U = 0` `:. q = -w= P_(ext) ( 10L -2L) = 0 (8) = 0` (i) ` q= - w = 1 atm xx 8 L = 8 L atm = 8 xx 101.3 J = 810.4 J ( 1L atm = 101. 3 J )` (ii) `q= -w= 2.303 RT log. (V_(2))/(V_(1)) = 2.303 PV log . (V_(2))/(V_(1)) = 2.303 xx 10 atm xx 2 L log. (10)/(2) = 32. 2 L atm.` `= 32.2 xx 101 . 3 J = 3261.86J`

Discussion

235.	Calculate the work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 500 ml to a volume of 2 L at 25°C and normal pressure.
Answer» n = 2 moles V_i = 500 ml = 0.5 lit V_f = 2 lit T = 25°C = 298 K w = -2.303 nRT log\((\frac{V}{V_i})\) w = -2.303 x 2 x 8.314 x 298 x log\(\frac{2}{0.5}\) w = -2.303 x 2 x 8.314 x 298 x log(4) w = -2.303 x 2 x 8.314 x 298 x 0.6021 w = -6871 J w =-6.871 kJ.

235.

Calculate the work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 500 ml to a volume of 2 L at 25°C and normal pressure.

Answer»

n = 2 moles

V_i = 500 ml = 0.5 lit

V_f = 2 lit

T = 25°C = 298 K

w = -2.303 nRT log\((\frac{V}{V_i})\)

w = -2.303 x 2 x 8.314 x 298 x log\(\frac{2}{0.5}\)

w = -2.303 x 2 x 8.314 x 298 x log(4)

w = -2.303 x 2 x 8.314 x 298 x 0.6021

w = -6871 J

w =-6.871 kJ.

Discussion

236.	If a gas at a pressure of `10 atm` at `300 K` expands against a constant external pressure of `2 atm` from a vol. `10` litres to `20` litres find work done ? [Isothermal process]
Answer» Process is irreversible `w= (-)underset(10)overset(20)int2dv= -2[20-10]=20 L. atm` `1` litre at `=101.3 J`

Discussion

237.	Four sample of ideal gas containig same moles and intially at same temperature and prcessure are subjected to four difference processes : (A) Isothermal reversible expansion (B) Isothermal irrversible expansion against final pressure (C ) Adiabatic reversible expansion (D) Adiabatic irreversible expansion against final pressure If in all the cases , final pressue is same then what will be the order of final temperture in the above cases.A. `T_(a)=T_(b)gt T_(c)=T_(d)`B. `T_(a)gtT_(b)gt T_(c)gtT_(d)`C. `T_(a)=T_(b)lt T_(c)ltT_(d)`D. `T_(a)=T_(b)gt T_(d)gtT_(c)`
Answer» Correct Answer - D

Discussion

238.	The temperature `T_(1) and T_(2)` of heat reservoirs in the ideal carnot engine are `15000^(@)C` and `500^(@)C` respectivley. If `T_(1)` increases by `100^(@)C`, what will be the efficiency of the engine?A. `62%`B. `59%`C. `95%`D. `100%`
Answer» Correct Answer - B Here, `T_(1)= 1500^(@)C+100^(@)C= 1600^(@)C` `=1600+273= 1873K` `T_(2)= 500^(@)C= 500+273= 773K` As `eta=1 -(T_(2))/(T_(1))= 1- (773)/(1873)= (1100)/(1873)` `eta= (1100)/(1873)xx100%= 59%`

Discussion

239.	A reversible engine takes in heat from a reservoir of heat at `527^(@)C`, and gives out to the sink at `127^(@)C` How many calories per second must it take from the reservoir in order to produce useful mechanical work at the rate of `750 wat t`? `1 cal. = 4.2J`.
Answer» Correct Answer - `357.1 cal//s` Here, `T_(1)= 527^(@)C= 800K, T_(2)= 127^(@)C= 400K` `Q_(1)=?, W=750 "wa tt"= 750 "joule"//sec` Efficiency, `eta=1 - (T_(2))/(T_(1))=1 - (400)/(800)= 1/2` As `eta=W/(Q_(1)) :. Q_(1)=W/(eta)=(750//4.2)/(1//2)=(1500)/(4.2)` `=357.1 cal//s`

Discussion

240.	A Carnot engine takes in heat from a reservoir of heat at `427^(@)C`. How many calories of heat must it take from the reservoir in order to procuce useful mechanical work at the rate of `357 watt`?
Answer» Here, `T_(1)= 427^(@)C = 427+273` `=700K` `T_(2)= 77^(@)C= (77+273)K= 350K`, `Q_(1)=?, W=357 "watt"`, As `(Q_(2))/(Q_(1))= (T_(2))/(T_(1))` `:. (Q_(1)-W)/(Q_(1))= (350)/(700)= 1/2` `Q_(1)-W= (Q_(1))/2 or (Q_(1))/2=W`, `Q_(1)= 2W=2xx357= 714 "watt"` `=(714)/(4.2)cal//s= 170 cal//s`

Discussion

241.	The temperatures `T_(1) and T(2)` of two heat reservoirs in an ideal carnot engine are `1500^(@)C and 500^(@)C`. Which of these (a) increasing `T_(1) by 100^(@)C` or (b) decreasing `T_(2) by 100^(@)C` would result in greater improvement of the efficiency of the engine?
Answer» Here, `T_(1)= 1500^(@)C= (1500+273)K` `= 1773K` `T_(2)= 500^(@)C= 500+273= 773K` `eta = 1 -(T_(2))/(T_(1))= 1 -(773)/(1773)= (1000)/(1773)` (a) Increasing `T_(1) by 100^(@)C` `T_(1)= 1600+273= 1873K, T_(2)=773K` `eta_(1)= 1 -(T_(2))/(T_(1))= 1 -(773)/(1873)= (1100)/(1873)` (b) Decreasing `T_(2) by 100^(@)C` `T_(1)= 1500+273= 1773K` `T_(2)=(500-100)+273= 673K` `eta_(2)= 1 -(T_(2))/(T_(1))= 1 -(673)/(1773)= (1100)/(1773)` Clearly, `eta_(2) gt eta_(1)`. Therefore, (b) is better choice.

Discussion

242.	Which of the following are state functions : (i) Height of hill (ii) Distance travelled in climbing the hill (iii) Energy change in climbing the hill.
Answer» Correct Answer - (i) and (ii)

Discussion

243.	Consider `p`-`V` diagram for an ideal gas shown in figure. Out of the following diagrams, which figure represents the `T-p` diagram ? (i) A. (iv)B. (ii)C. (iii)D. (i)
Answer» Correct Answer - C According to the question given that `pV` = constant Hence, we can say that the gas is going through an isothermal process. Clearly, from the graph that between process 1 and 2 themperature is constant and the gas expands and pressure decreases i.e., `p_(2)ltp_(1)` which corresponds to diagram (iii).

Discussion

244.	If an average jogs, he produces `14.5xx10^(3)` cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires `580xx10^(3)` cal for evaporation) isA. 0.25 KgB. 2.25 KgC. 0.05 kgD. 0.20 Kg
Answer» Correct Answer - A Amount of sweat evaporated per minute `=("calories produed per mimute")/("no of calories required for evaporation per kg")` `=(14.5xx10^(4))/(580xx10^(3))` =0.25 Kg

Discussion

245.	An ideal gas undergoes cyclic process ABCDA as shown in given P-V diagram (Fig. 12.4). The amount of work done by the gas is(a) 6P0V0(b) –2 P0V0(c) + 2 P0V0(d) + 4 P0V0
Answer» The amount of work done by the gas is (b) –2 P₀V₀

Discussion

246.	One mole of an ideal gas undergoes a cyclic change ABCDA as shown in (figure). What is the net work done (in joule) in the process? Take `1 atm = 10^(5)Pa`.
Answer» Correct Answer - 4 Work done= area ABCDA `= ABxxBC` `=[(0.3-0.1)xx10^(-3)]xx(0.4-0.2)10^(5)= 4J`

Discussion

247.	An ideal gas of mass `m` in a state `A` goes to another state `B` via three different processes as shown in Fig. If `Q_(1), Q_(2)` and `Q_(3)` denote the heat absorbed by the gas along the three paths, then A. Change in internal energy in all the three paths is equal.B. In all the three paths heat is absorbed by the gas.C. Heat absorbed/released by the gas is maximum in path (1).D. Temperature ot the gas first increases and then decreases continously in path (1).
Answer» Correct Answer - A::B::C Internal energy `(U)` depends only on the initail and final state.Hence, `DeltaU` will be same in all the three paths. In all the three paths. In all the three paths work done by the gas is positive and the product `PV` or temperature `T` is incerasing. Therefore, internal energy is also increasing. So, from the first law of thermodynamics, heat will be absorbed by the gas. Further, area under `P-V` graph is maximum in path `1` while `DeltaU` is same for all the three paths Therefore, heat absobed by the gas, is maximum in path`1`. For temperature of the gas, wecan see that product `PV` first increases in path `1` but whether in is decreasing or increasing later on we cannot say anything about it unless the exact values are known to us.

Discussion

248.	A tyre pumped to a pressure of `6 atmosphere` bursts suddenly. Calculate the temperature of escaping air. Given initial room temperature is `15^(@)C` and gamma for air is 1.4`.
Answer» Correct Answer - `176.2K` `P_(1)= 6 atm. P_(2)= 1 atm` `T_(1)= 273+15= 288K, T_(2)= ? gamma= 1.4` From `P_(2)^((1-gamma))T_(2)^(gamma)= P_(1)^((1-gamma))T_(1)^(gamma)` `T_(2)= T_(1)((P_(1))/(P_(2)))^((1-gamma)//(gamma))=288(6/1)^(((1-1.4))/(1.4))` `=288xx6^(-2//7)= 172.6K`

Discussion

249.	One mole of an ideal gas is taken through the cyclic process ABCDA, as shown in (figure). Using the graph, calculate(i) Work done in the processes `ArarrB, BrarrC, CrarrD and DrarrA` (ii) Work done in complete cycle ABCDA (iii) Heat rejected by the gas in one complete cycle.
Answer» (i) In process A to B, volume is constant `:. dW_(AB)= P.dV= Zero` In process BC `dW_(BC)= area BEFC =BExxBC` `=(5xx1.013xx10^(5)N//m^(2)xx50)J` `=2.53xx10^(7)J` In process CD, volume remains constant `:. dW_(CD)= P.dV=Zero` In process DA `dW_(DA)= area AEFD= AExxEF` `=(10xx1.013xx10^(5))xx(-50)` `= -5.06xx10^(7)J`. (ii) Net work done in complete cycle ABCDA `dW=dW_(AB)+dW_(BC)+dW_(CD)+dW_(DA)` `= 0+2.53xx10^(7)+0-5.06xxc10^(7)` `= -2.53xx10^(7)J` (iii) As the process is cyclic, `dU=0`. From first law of thermodynamics `dQ=dU+dW` `= 0-2.53xx10^(7)= -2.53xx10^(7)J` Negative sign shows that this much heat is rejected in one complete cycle.

Discussion

250.	Can a system be heated and its temperature remains constant?
Answer» Yes, a system can be heated and its temperature remains constant. This would happen when the entire heat supplied to the system is spent up in expansion (i.e.,Working against the surroundings).

Discussion

Explore topic-wise InterviewSolutions in .

If a gas of volume 10 L is expanded to quadruple its volume at 4 atm pressure, the external work done isA. 3 kJB. 6 kJC. 12 kJD. 18 kJ

5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be `T_1,` the work done in the process isA. `(9)/(8)RT_(1)`B. `(3)/(2)RT_(1)`C. `(15)/(8)RT_(1)`D. `(9)/(2)RT_(1)`

A sample of ideal gas `(gamma=1.4)` is heated at constant pressure. If an amount of 100 J heat is supplied to the gas, the work done by the gas isA. 28.57 JB. 56.54 JC. 38.92 JD. 65.38 J

The amount of heat supplied to `4xx10^(-2)` kg of nitrogen at room temperature to rise its temperature by `50^(@)C` at constant pressure is (Molecualr mass of nitrogen is 28 and `R= 8.3 J mol^(-1)K^(-1))`A. 2.08 KJB. 3.08 KJC. 4.08 KJD. 5.08 KJ

`1672cal` of heat is given to one mole of oxygen at `0^(@)C` keeping the volume constant. Raise in temperature is `(c_(p)=0cal//gm^(0)k and R=2cal//"mole"//K)`A. `33.6^(@)C`B. `36.3^(@)C`C. `63.3^(@)C`D. `334.4^(@)C`

`0.5` mole of diatomic gas at `27^(@)C` is heated at constant pressure so that its tripled. If `=8.3J "mole"^(-1)k^(-1)` then work done isA. `4980J`B. `2490J`C. `630J`D. `1345J`

Two metal balls having masses `50g` and `100g` collides with a target with same velocity. Then the ratio of their rise in temperature isA. `1:2`B. `4:1`C. `2:1`D. `1:1`

A gas is compressed adiabatically till its temperature is doubled. The ratio of its final volume to initial volume will beA. `1/2`B. more then `1/2`C. less then `1/2`D. Between 1 and 2

A tyre filled with air,`(27^(@)C`,o and 2 atm ) bursts, then what is temperature of air `(lambda=1.5)`A. `-33^(@)C`B. `0^(@)C`C. `27^(@)C`D. `240^(@)C`

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 = 2RTB. Q = RTC. Q = 3RTD. Q = 4RT

In a thermodynamic process on an ideal diatomic gas, work done by the gas is `eta` times. The heat supplied `(eta lt 1)`. The molar heat capacity of the gas for the process isA. `(5R)/(2)`B. `5eta(R)/(2)`C. `(5R)/(2(1-eta))`D. `(5R(1-eta))/(2)`

When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied, which increases the internal energy of the gas, isA. `2/5`B. `3/5`C. `3/7`D. `5/7`

One mole of an ideal gas at `27^@C` expanded isothermally from an initial volume of `1 ` litre to `10` litre. The `DeltaU` for this process is : `(R=2 cal K^(-1) mol^(-1))`A. zeroB. `9` litre-atmC. `1281.1 cal`D. `164.7 cal`

Enerrgy can transfter from syetm to surroundings as work ifA. There is thermal equilibrium between system and surrounding.B. There is mechanical equilibrium between system and surrounding.C. If pressute of system gt atmospheric pressure.D. None of these.

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 A. 3B. `5/2`C. `5/3`D. `7/2`

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 A. `3`B. `5//2`C. `5//3`D. `7//2`

With what pressure must a given volume of oxygen, originally at 300K and 1 bar pressure be adiabatically and irreversibly compored in order to raise its temperture to 600K `(gamma=7//5)`A. `4.5 "bar"`B. `9"bar"`C. `2.25"bar"`D. `6.75"bar"`

2 mole of ideal gas expands isothermically and reversibally from 1 L to 10 L at 300 K. then `DeltaH` is :A. 4.98 kJB. 11.47 kJC. `-11.47 kJ`D. zero

`1dm^(3)` of an ideal gas at a pressure of 10bar expands reversibly and isothermally to final volume of 10 litre .(In 10=2.3). Heat absorbed in the process will be:A. `1.15kJ`B. `4.6J`C. `2.3kJ`D. `9.2kJ`

For a chemical reaction the values of ∆H and ∆S at 300 K are – 10 kJ mol-1 and -20 J mol-1 respectively. What is the value of AG of the reaction? Calculate the ∆G of a reaction at,600K assuming ∆H and AS values are constant. Predict the nature of the reaction.

Calculate the work done (in joules) when 0.2 mole of an ideal gas at 300 K expands isothermally and reversibly from initial volume of 2.5 litres to the final volume of 25 litres.A. 996B. 1148C. 11.48D. 897

Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?

Two litres of an ideal gas at a pressure of 10 atm expands isothermally at `25^(@)C` into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?

Two litres of an ideal gas at a pressure of 10 atm expands isothermally at `25^(@)C` into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?A. 10 J, 10 JB. 8 J, 10 JC. 18 J, 0 JD. 0 J, 0 J

What is the entropy change when 1 mole oxygen gas expands isothermally and reversibly from an initial volume of 10 L to 100 L at 300 K ?A. `"19.14 J K"^(-1)`B. `"109.12 J K"^(-1)`C. `"29.12 J K"^(-1)`D. `"10 J K"^(-1)`

Calculate the work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 500 ml to a volume of 2 L at 25°C and normal pressure.

If a gas at a pressure of `10 atm` at `300 K` expands against a constant external pressure of `2 atm` from a vol. `10` litres to `20` litres find work done ? [Isothermal process]

The temperature `T_(1) and T_(2)` of heat reservoirs in the ideal carnot engine are `15000^(@)C` and `500^(@)C` respectivley. If `T_(1)` increases by `100^(@)C`, what will be the efficiency of the engine?A. `62%`B. `59%`C. `95%`D. `100%`

A reversible engine takes in heat from a reservoir of heat at `527^(@)C`, and gives out to the sink at `127^(@)C` How many calories per second must it take from the reservoir in order to produce useful mechanical work at the rate of `750 wat t`? `1 cal. = 4.2J`.

A Carnot engine takes in heat from a reservoir of heat at `427^(@)C`. How many calories of heat must it take from the reservoir in order to procuce useful mechanical work at the rate of `357 watt`?

The temperatures `T_(1) and T(2)` of two heat reservoirs in an ideal carnot engine are `1500^(@)C and 500^(@)C`. Which of these (a) increasing `T_(1) by 100^(@)C` or (b) decreasing `T_(2) by 100^(@)C` would result in greater improvement of the efficiency of the engine?

Which of the following are state functions : (i) Height of hill (ii) Distance travelled in climbing the hill (iii) Energy change in climbing the hill.

Consider `p`-`V` diagram for an ideal gas shown in figure. Out of the following diagrams, which figure represents the `T-p` diagram ? (i) A. (iv)B. (ii)C. (iii)D. (i)

If an average jogs, he produces `14.5xx10^(3)` cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires `580xx10^(3)` cal for evaporation) isA. 0.25 KgB. 2.25 KgC. 0.05 kgD. 0.20 Kg

An ideal gas undergoes cyclic process ABCDA as shown in given P-V diagram (Fig. 12.4). The amount of work done by the gas is(a) 6P0V0(b) –2 P0V0(c) + 2 P0V0(d) + 4 P0V0

One mole of an ideal gas undergoes a cyclic change ABCDA as shown in (figure). What is the net work done (in joule) in the process? Take `1 atm = 10^(5)Pa`.

A tyre pumped to a pressure of `6 atmosphere` bursts suddenly. Calculate the temperature of escaping air. Given initial room temperature is `15^(@)C` and gamma for air is 1.4`.

One mole of an ideal gas is taken through the cyclic process ABCDA, as shown in (figure). Using the graph, calculate(i) Work done in the processes `ArarrB, BrarrC, CrarrD and DrarrA` (ii) Work done in complete cycle ABCDA (iii) Heat rejected by the gas in one complete cycle.

Can a system be heated and its temperature remains constant?