4131 + Interview Questions in THERMODYNAMICS IN GENERAL AWARENESS Page 61 InterviewSolution

3001.	The vapour pressure of benzene is `1.53 xx 10^(4) Nm^(-2)` at 303 K and `5.2 xx 10^(4) Nm^(-2)` at 333 K. Calculate the mean latent heat of evaportion of benzene over this temprature range
Answer» Correct Answer - 31.1 kj Using clausinus-Clapeyron equation, 2.303 log `(P_(2))/(P_(1)) = (Delta H_(v))/(R) ([T_(2) - T_(1)])/(T_(1)T_(2))` `Delta H_(v) = (2.303 RT_(1) T_(2))/((T_(2) - T_(1))) log. (P_(2))/(P_(1))` We have `Delta H_(V) = (2.303 xx 8.314 xx 303 xx 333)/((333 - 300)) log_("to"). (5.2 xx 10^(4))/(1.53 xx 10^(4))` `=31.1 xx 10^(3) J = 31.1 kJ`

Discussion

3002.	State the third law of thermodynamics.
Answer» It states that the entropy of pure crystalline substance at absolute zero is zero. (or) It can be stated as “it is impossible to lower the temperature of an object to absolute zero in a finite number of steps”. Mathematically \(\lim\limits_{T \to 0} S = 0\)

3002.

State the third law of thermodynamics.

Answer»

It states that the entropy of pure crystalline substance at absolute zero is zero.

(or)

It can be stated as “it is impossible to lower the temperature of an object to absolute zero in a finite number of steps”. Mathematically \(\lim\limits_{T \to 0} S = 0\)

Discussion

3003.	Assertion : The ratio `C_(P)//C_(upsilon)` is more for helium gas than for hydrogen gas. Reason : Atomic mass of helium is more than that of hydrogen.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true
Answer» Correct Answer - B

Discussion

3004.	Consider cyclic process as depicted in figure. Match column-I and column-II for questions. Q. Match the following .
Answer» Correct Answer - A::B::C::D

Discussion

3005.	Consider cyclic process as depicted in figure. Match column-I and column-II for questions. Q. Match the following
Answer» Correct Answer - A::B::C

Discussion

3006.	Match the following
Answer» Correct Answer - A::B::C::D

Discussion

3007.	Match the following
Answer» Correct Answer - A::B::C::D

Discussion

3008.	Match the following
Answer» Correct Answer - A::B::C::D

Discussion

3009.	The variation of pressure versus volume is shown in the figure. The gas is diatomic and the molar specific heat capacity for the process is found to xR. Find the value of x. .
Answer» Correct Answer - 3

Discussion

3010.	2 moles of a diatomic gas are enclosed in a cylinder piston arrangment. The area of cross section and mass of the piston are `1 cm^(2)` and 1 kg respectively. A heater is supplying heat to the gas very slowly. Find heat supplied (in joule) by the heater is the piston moves through a distance of 10 cm. .
Answer» Correct Answer - 7

Discussion

3011.	One mole of an ideal gas is taken from a to b along two parths dented by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is `W_(s) ` and that along the dotted line path is `W_(d)` then the interger closet to the ratio `W_(d) // W_(s)` is
Answer» Correct Answer - -2

Discussion

3012.	Making use of the result obtained in the foregoing problem. Find the what temperature the isothermal compressibility `x` of a Van der Walls gas is greater than that of an ideal gas. Examine the case when the molar volume is much greater than the parameter `b`.
Answer» For an ideal gas `K_0 = (V)/(RT)` Now `K = ((V - b)^2)/(RTV){1 - (2a(V - b)^2)/(RTV^3)}^-1 = K_0(1 - (b)/(V))^2{1 - (2a)/(RTV)(1 - (b)/(V))^2}^-1` =`K_0{1 - (2b)/(V)+(2a)/(RTV)}`, to leading order in `a, b` Now `K gt K_0`if `(2 a)/(RTV) gt (2 b)/(V)` or `T lt (a)/(bR)` If `a,b` do not vary much with temperature, then the effect at high temperature is clearly determined by `b` and its effect is repulsive so compressibility is less.

Discussion

3013.	Find the isothermal compressibility `x` of a Van der Walls gas as a function of volume `V` at temperature `T` By definition, `x = - (1)/(V) (del V)/(del p)`.
Answer» `p =(R T)/(V - b) =(a)/(V^2) - V((del p)/(del V))_T = (R TV)/((V - b)^2)-(2a)/(V^2)` or, `K=(-1)/(V)((del V)/(del p))_T` =`[(RTV^3 - 2a(V - b)^2)/(V^2(V - b)^2)]^-1 = (V^2(V - b))/[[RTV^3 - 2a(V -b)^2]`.

Discussion

3014.	One mole of a certain gas is contained in a vessel of volume `V = 0.250 1`. At a temperature `T_1 = 300 K` the gas pressure is `p_1 = 90 atm`, and at a temperature `T_2 = 350 K` the pressure is `p_2 = 110 atm`. Find the Van der Walls parameters for this gas
Answer» `p_1 = RT_1 (1)/(V^2), p_2 = R T_2 (1)/(V - b) - (a)/(V^2)` So, `p_2 - p_1 = (R(T_2 - T_1))/(V - b)` or, `V - b = (R(T_2 - T_1))/(p_2 - p_1)` or , `b = V - (R(T_2 - T_1))/(p_2 - p_1)` Also, `p_1 = T_1 (p_2 - p_1)/(T_2 - T_1) -(a)/(V^2)`, `(a)/(V^2) = (T_1(p_1 - p_2))/(T_2 - T_1) - p_1 = (T_1 p_2 - p_1 T_2)/(T_2 - T_1)` or, `a = V^2 (T_1 p_2 - p_1T_2)/(T_2 - T_1)` Using `T_1 = 300 K, p_1 = 90 atms, T_2 = 350 K, p_2 = 110 atm, V = 0.250 "litre"` `a = 1.87 "atm.litre"^2//"mole"^2, b = 0.045 litre//"mole"`.

Discussion

3015.	One mole of nitrogen is contained in a vessel of volume `V = 1.00 1`. Find : (a) the temperature of the nitrogen at which the pressure can be calculated from an ideal gas law with an error `eta = 10 %` (as compared with the pressure calculated from the Van der Walls equation of state) , (b) the gas pressure at this temperature.
Answer» (a) `p = [(RT)/(V_m - b) -(a)/(V_M^2)](1 + eta) = (RT)/(V_M)` (The pressure is less for a Vander Wall gas than for an ideal gas) or, `(a(1 + eta))/(V_M^2) = RT[(-1)/(V_M) + (1 + eta)/(V_M - b)] = RT (eta V_M + b)/(V_M(V_M - b))` or, `T = (1(1 + eta)(V_M - b))/(R V_M(eta V_M + b))`, (here `V_M` is the molar volume) =`(1.35 xx 1.1 xx (1-0.039))/(0.082 xx (0.139))~~ 125 K` (b) The corresponding pressure is `p =(RT)/(V_M - b) -(a)/(V_M^2) =(a(1 + eta))/(V_M(eta V_M + b))- (a)/(V_M^2)` =`(a)/(V_M^2) ((V_M + eta V_M - eta V_M - b))/((eta V_M + b)) = (a)/(V_M^2) ((V_M - b))/((V_M + b))` =`(1.35)/(1) = (0.961)/(0.139) ~~ 9.2 atm`.

Discussion

3016.	Under what pressure will carbon dioxide have the density `rho = 500 g//1` at the temperature `T = 300 K` ? Carry out the calculations both for an ideal and for a Van der Walls gas.
Answer» For an ideal gas law `p = (rho)/(M) R T` So, `p = 0.082 xx 300 xx (500)/(44) atms = 279.5` atmosphere For Vander Wall gas Eqs. `(p + (v^2 a)/(V^2))(V - vb) = vRT`, where `V = vV_M` or, `p = (v RT)/(V - vb) = (av^2)/(V^2) = (m RT//M)/(V -(mb)/(M)) -(am^2)/(V^2 M^2)` =`(rho RT)/(M - rho b) -(a rho^2)/(M^2) = 79.2 atm`.

Discussion

3017.	An ideal gas of molar mass `M` is located in the uniform gravitational field in which the free-fall acceleration is equal to `g`. Find the gas pressure as a function of height `h`. If `p = p_0` at `h = 0`, and the temperature varies with height as (a) `T = T_0 (1 - ah)` , (b) `T = T_0 (1 + ah)`, where `a` is a positive constant.
Answer» (a) We know that the variation of pressure with height of a fluid is given by : `dp = - rho g dh` But from gas law `p = (rho)/(M) RT` or, `rho = (p M)/(RT)` From these two Eqs. `dp = - (p Mg)/(RT) dh` ….(1) ltrgt or, `(dp)/(p) = (-Mg dh)/(R T_0(1 - ah))` Integrating, `int_(p_0)^p (dp)/(p) = (-Mg)/(R T_0) int_0^h (dh)/((1 - ah))`, we get `1 n (p)/(p_0) = 1 n(1 - ah)^(Mg//aRT_0)` Hence, `p = p_0(1 - ah)^(Mg//aRT_0)`, Obvionsly `h lt (1)/(a)` (b) Proceed up to Eq. (1) of part (a). and then put `T = T_0 (1 + ah)` and proceed further in the same fashion to get `p = (p_0)/((1 + ah)^(Mg//aRT_0))`.

Discussion

3018.	An ideal gas of molar mass `M` is contained in a very tall vertical cylindrical vessel in the uniform gravitational field in which the free-fall acceleration equals to `T`, find the height at which the centre of gravity of the gas is located.
Answer» As the gravitational field is constant the centre of gravity and the centre of mass are same. The location of `C.M.` `h = (int_0^oo h dm)/(int_0^oo dm) = (int_0^oo h rho dh)/(int_0^oo rho dh)` But from Barametric formula and gas law `rho = rho_0 e^(-Mg h//RT)` So, `h = (int_0^alpha h(e^(-Mg h//RT))dh)/(int_0^alpha (e^(-Mg h//RT))dh) = (RT)/(Mg)`.

Discussion

3019.	An ideal gas of molar mass `M` is contained in a tall vertical cylindrical vessel whose base area is `S` and height `h`. The temperatuer of the gas is `T`, its pressure on the bottom bass is `p_0`. Assuming the temperature and the free-fall acceleration `g` to be independent of the height, find the mass of gas in the vessel.
Answer» From the Barametric formula, we have `p = p_0 e^(-Mg h//RT)` and from gas law `p = (p M)/(RT)` So, at constant temperature from these two Eqs. `rho = (M p_0)/(RT) e^(-Mg h//RT) = rho_0 e^(-Mg h//Rt)` ….(1) eq.(1) shows that density varies with height in the same manner as pressure. Let us consider the mass element of the gas contained in the column. `dm = rho (S dh) = (M p_0)/(RT) e^(-Mg h//Rt) S dh` Hence the sought mass, `m = (M p_0 S)/(RT) int_0^h e^(-Mg h//RT) dh = (p_0 S)/(g) (1 - e^(-Mg h//RT))`.

Discussion

3020.	A chamber of volume `V = 87 1` is evacuated by a pump whose evacuation rate (see Note the foregoing problem) equals `C = 10 1//s`. How soon will the pressure in the chamber decrease by `eta = 1000` times ?
Answer» Let `rho` be the instantaneous density, then instantaneous mass `= V_p`. In a short interval `dt` the volume is increased by `Cdt`. So, `V_rho = (V + Cdt) (rho + d rho)` (because mass remains constant in a short interval `dt`) so `(d rho)/(rho) = - (C)/(V) dt` Since pressure `alpha` density `(dp)/(p) = - (C)/(V) dt` or `int_(p_1)^(p_2) - (dp)/(p) = (C)/(V) t`, or `t = (V)/(C) 1n (p_1)/(p_2) = (V)/(C) 1n (1)/(eta) 1.0 min`.

Discussion

3021.	Assuming the temperature and the molar mass of air, as well as the free-fall acceleration, to be independent of the height, find the difference in heights at which the air densities at the temperature `0 ^@C` differ. (a) `e` times , (b) by `eta = 1.0 %`.
Answer» We have `dp = - rho g dh` but from gas law `p = (rho)/(M) RT`, Thus `dp = (d rho)/(M) RT` at const. temperature So, `(d rho)/(rho) = (g M)/(RT) gh` Integrating within limits `int_(rho_0)^rho (d rho)/(rho) = int_0^h (g M)/(RT) gh` or, `1 n (rho)/(rho_0) = -(g M)/(RT) h` So, `rho = rho_0 e^(-M gh//RT)` and `h = - (RT)/(M g) 1 n (rho)/(rho_0)` (a) Given `T = 273^@K, (rho_0)/(rho) = e` Thus `h = - (RT)/(M g) 1 n e^-1 = 8 km`. (b) `T = 273^@ K` and `(rho_0 - rho)/(rho_0) = 0.01` or `(rho)/(rho_0) = 0.99` Thus `h = - (RT)/(M g) 1 n(rho)/(rho_0) = 0.09 km` on substitution.

Discussion

3022.	Suppose the pressure `p` and the density `rho` of air are related as `p//rho^n = const` regardless of height (`n` is a constant here). Find the corresponding temperature gradient.
Answer» We have, `(dp)/(dh) = - rho g` (Sec 2.13)…(1) But, from `p = C rho^n` (where `C` is, a const) `(dp)/(d rho) = Cn rho^(n - 1)` ….(2) We have from gas low `p = rho (R )/(M) T`, so using (2) `C rho^n = rho ( R)/(M).T`, or `T = (M)/(R) C rho^(n - 1)` Thus, `(dT)/(d rho) = (M)/(R) .C (n - 1) rho^(n - 2)` ...(3) But, `(d T)/(dh) = (dT)/(d rho).(d rho)/(dp).(dp)/(dh)` So, `(dT)/(dh) = (M)/(R) C(n - 1) rho^(n -1) (1)/(C n rho^(n -1))(- rho g)= (- Mg (n -1))/(n R)`.

Discussion

3023.	Find the minimum attainable pressure of ideal gas in the process `T = T_0 + alpha V^2`, where `T_0` and `alpha` positive constants, and `V` is the volume of one mole of gas. Draw the approximate `p vs V` plot of this process.
Answer» `T = T_0 + alpha V^2 = T_0 + alpha (R^2 T^2)/(p^2)` (as, `V = RT//p-` for one mole of gas) So, `p = sqrt(alpha) RT (T - T_0)^(1//2)` ….(1) For `P_(min), (dp)/(dT) = 0`, which gives `T = 2 T_0` …(2) From (1) and (2), we get, `p_(min) = sqrt(alpha) R 2 T_0 (2 T_0 - T_0)^(-1//2) = 2 R sqrt(alpha T_0)`.

Discussion

3024.	In view of the signs of `Delta_(r)G^(0)` for the following reactions `PbO_(2)+Pbrarr2PbO,Delta_(r)G^(0)lt0` `SnO_(2)+Snrarr2SnO,Delta_(r)G^(0)gt0` Which oxidation state are more characteristic for lead and tin?A. For lead +4, for tin +2B. For lead +2, for tin +2C. For lead +4, for tin +4D. For lead +2, for tin +4
Answer» Correct Answer - D `{:(,PbO_(2),+,Pb,rarr,2PbO),("Oxidation state",+4,,0,,+2):}` Since, `DeltaG^(@) lt 0`, i.e., it is negative. therefore, the reaction is spontaneous in the forward direction. This suggest that `Pb^(2+)` is more stable than `Pb^(4+)` `{:(,SnO_(2),+,Sn,rarr,2SnO),("Oxidation state",+4,,0,,+2):}` Since, `DeltaG^(@) lt 0` i.e., it is positive. Therefore, the reaction is non-spontaneous in the forward direction. but it will spontaneous in the backward direction. This suggests that `Sn^(2+)` is less stable than `Sn^(4+)`. These facts are also supported by the inert pair effect down the group.

Discussion

3025.	The incorrect expression among the following isA. `(DeltaG_("system"))/(DeltaS_("total"))=-T`B. In isothermal process, `w_("reversible")=-nRT "In" (V_(f))/(V_(i))`C. Ink `= (DeltaH^(@)-TDeltaS^(@))/(RT)`D. `K=e^(-DeltaG^(@)//RT)`
Answer» Correct Answer - C `DeltaG^(@) =DeltaH^(@) - TDeltaS^(@)` `-RTInK = DeltaH^(@) - TDeltaS^(@)` Ink`=-(DeltaH^(@) - TDeltaS^(@))/(RT)`

Discussion

3026.	Among them intensive property isA. MassB. VolumeC. Surface tensionD. Enthalpy
Answer» Correct Answer - C

Discussion

3027.	The heat or rectin does not depend upon :A. Number of steps in which the reaction is carried outB. Temperature at which the reaction is carrired outC. physical state of reactant and productsD. Whether the reaction the reaction is carried out at constant pressure or at constant volume
Answer» Correct Answer - a

Discussion

3028.	Which of the following enthalpy may be positive or negative?A. Enthalpy of atomisationB. Enthalpy of combustionC. Enthalpy of solutionD. Enthalpy of hydration
Answer» Correct Answer - c

Discussion

3029.	The following reaction is performed at 298K `2NO(g) + O_(2)(g) hArr 2NO_(2)(g)` The standard free energy of formation of NO(g) is 86.6 KJ/mol at 298 K. What is the standard free energy formation of `NO_(2)(g) "at" 298`? `(K_(p) = 1.6 xx 10^(12))`A. R(298) in `(1.6 xx 10^(12))` -86600B. 86600 + R(298) In `(1.6xx 10^(12))`C. `86600 -("In" (1.6 xx 10^(12)))/(R(298))`D. `0.5 [2 xx 86,600 -R(298) "In" (1.6 xx 10^(12))]`
Answer» Correct Answer - D `2DeltaG_(f(NO_(2)))^(@) -[2DeltaG_(f(NO))^(@)+DeltaG_(f(O_(2)))^(@)] = DeltaG_(r)^(@) =- RTlnK_(p)` `2DeltaG_(f(NO_(2))^(@) -[2xx 86,600+0] =- RTlnK_(p)` `DeltaG_(f(NO_(2))^(@) = 0.5[2xx86,600-R(298) ln(1.6 xx 10^(12))]`

Discussion

3030.	What is the equilibrium constant `K_(c)` for the following reaction at `400K`? `2NOCI(g) hArr 2NO(g) +CI_(2)(g)` `DeltaH^(Theta) = 77.2 kJ mol^(-1)` and `DeltaS^(Theta) = 122 J K^(-1) mol^(-1) at 400K`.
Answer» According to Gibbs-Helmhotz equation `DeltaG^(Theta) = DeltaH^(Theta) - T DletaS^(Theta)` `= 77.2 - 400 xx 122 xx 10^(-3)` `28.4 kJ = 28.4 xx 10^(3)J` We know `DeltaG^(Theta) =- 2.303 RT log_(10)K_(c)` `:. log_(10)K_(c) = (-DeltaG^(@))/(2.303RT)` `=(28.4xx10^(3))/(2.303xx8.314xx400) = - 3.7081` `K_(c) = antilog (-3.7081) = 1.958 xx 10^(-4)`

Discussion

3031.	The reaction, `4Ag(s)+O_(2)(g) to 2Ag_(2)O(s), "is exothermic"` Which statement about the reaction is correct?A. It is spontaneous at all temperatureB. It is spontaneous only at low temperaturesC. It is spontaneous only at high temperatures.D. It is non-spontaneous at all temperatures.
Answer» Correct Answer - B

Discussion

3032.	The equilibrium constant for the reaction `CO_(2)(g) +H_(2)(g) hArr CO(g) +H_(2)O(g) at 298 K` is `73`. Calculate the value of the standard free enegry change `(R =8.314 J K^(-1)mol^(-1))`
Answer» We know, `DeltaG^(Theta) =- 2.303 RT log K_(c)` Given, `K_(c) = 70, R = 8.314 J K^(-1) mol^(-1), T = 298 K` Therefore, form the above equation `DeltaG^(Theta) = - 2.303 xx 8.314 xx 298 log_(10)70 =- 10527 kJ`

Discussion

3033.	The equilibrium constant at `25^(@)C` for the process: `CO^(3+) (aq) +6NH_(3)(aq) hArr[Co(NH_(3))_(6)]^(3+)(aq)` is `2 xx 10^(7)`. Calculate the value of `DeltaG^(Theta)` at `25^(@)C at 25^(@)C[R = 8.314 J K^(-1)mol^(-1)]`. In which direction the reaction is spontaneous when the recatants and proudcts are in standard state?
Answer» We know, `DeltaG^(Theta) =- 2.303 RT log_(10)K_(c)` Given, `K_(c) =2xx10^(5),T = 298 K,R = 8.314 J K^(-1)mol^(-1)` Thus, form the above equation, `DeltaG^(Theta) =- 2.303 xx 8.314 xx 298 log 2xx10^(5) = 8.588 kJ`

Discussion

3034.	`A` man of `60` kg gains `1000` cal of heat by eating `5` mangoes. His efficiency is `29%`. To what height he can jump by using this energy?A. `2m`B. `20 m`C. `28 m`D. `0.2 m`
Answer» Correct Answer - A `"Efficiency" `xx` "energy=work done=mgh"`

Discussion

3035.	Calculate the work performed when `2` moles of hydrogen expand isothermally and reversibly at `25^(@)C` from `15` to `50` litres. (in cal)
Answer» We have, `W= -2.303n RT "log"(V_(2))/(V_(1))= -2.303xx2xx2xx298xx"log"(50)/(15)= -1436` calories

Discussion

3036.	What are the signs of ` DeltaH_(r)and DeltaS_(r) "for the reaction"` `2C(s)+O_(2)(g)to2CO(g)`A. `{:(,DeltaH,DeltaS,,),(,-,-,,):}`B. `{:(,DeltaH,DeltaS,,),(,-,+,,):}`C. `{:(,DeltaH,DeltaS,,),(,+,+,,):}`D. `{:(,DeltaH,DeltaS,,),(,+,-,,):}`
Answer» Correct Answer - B

Discussion

3037.	Calcualte the enthalpy change when infinitely dilute solution of `CaCI_(2)` and `Na_(2)CO_(3)` are mixed. `Delta_(f)H^(Theta)` for `Ca^(2+)(aq), CO_(3)^(2-)(aq)`, and `CaCO_(3)(s)` are `-129.80, -161.65`, and `-288.50 kcal mol^(-1)` respectively.
Answer» Infinitely dilute solutions are completely ionised. `{:(Caunderset(darr).^(2+)CI_(2)^(2-) +Na_(2)^(2+)CO_(3)^(2-)),(CaCO_(3)darr+2NaCI):}` `{:(Caunderset(darr).^(2+)+2CI^(Theta)+2Na^(darr)+CO_(3)^(2-)),(Ca^(2+)+CO_(3)^(2-)+2Na^(o+)+CO_(3)^(2-)):}` `Ca^(2+) +CO_(3)^(2-) rarr CaCO_(3)(s)` `-DeltaH^(Theta) = Delta_(f)H^(Theta)CaCO_(3) -[Delta_(f)H^(Theta)ca^(2+)+Delta_(f)H^(Theta)CO_(3)^(2-)]` `= - 288.5 -(-129.80-161.65)` `DeltaH^(Theta) = 2.95 kcal`

Discussion

3038.	Which of the following equation gives the values of heat of formation `(DeltaH_(f)^(@))`?A. `C("diamond")+O_(2)(g)toCO_(2)(g)`B. `(1)/(2)H_(2)(g)+(1)/(2)F_(2)(g)toHF(g)`C. `N_(2)(g)+3H_(2)(g)to2NH_(3)(g)`D. `H_(2)(g)+F_(2)(g)to2HF(g)`
Answer» Correct Answer - B

Discussion

3039.	For reduction of ferric oxide by hydrogen , `Fe_(2)O_(3)(s)+3H_(2)(g)rarr2Fe(s)+3H_(2)O(l)`. `Delta H_(300)^(o)= 26.72 kJ`. The reaction was found to be too exothermic. To be convenient, it is desirable that `DeltaH^(0)` should be at the most `-26 kJ`. At what temperature difference it is possible ? `C_(p)[FeO_(3)]=105, C_(p)[Fe(s)]=25, C_(p)[H_(2)O(l)]=75, C_(p)[H_(2)(g)]=30` (all are in J/mol)
Answer» Correct Answer - 9

Discussion

3040.	Refer above figure and answer the following questions. (i) What is the work done in process AB ? (ii) What is the change in internal energy and heat released in process BC ?
Answer» (i) In this case, the change in the internal energy is zero, as the temperature of the gas remains constant. Hence, the work done, W = heat absorbed, Q_H . (ii) In this case, the change in the internal energy, ∆ U = _nC_V(T_C – T_H), where n = number of moles of the gas used in the Stirling engine and C_V = molar specific heat of the gas. As W = 0 at constant volume, heat released’= ∆ U.

3040.

Refer above figure and answer the following questions. (i) What is the work done in process AB ? (ii) What is the change in internal energy and heat released in process BC ?

Answer»

(i) In this case, the change in the internal energy is zero, as the temperature of the gas remains constant. Hence, the work done, W = heat absorbed, Q_H .

(ii) In this case, the change in the internal energy, ∆ U = _nC_V(T_C – T_H), where n = number of moles of the gas used in the Stirling engine and C_V = molar specific heat of the gas. As W = 0 at constant volume, heat released’= ∆ U.

Discussion

3041.	If a Carnot refrigerator works between 250 K and 300 K, its coefficient of performance = (A) 6 (B) 1.2 (C) 5 (D) 10.
Answer» Correct option is (C) 5

Discussion

3042.	Draw a neat labelled diagram of a Sterling cycle and describe the various processes taking place in a Sterling engine.Sterling cycle P - V diagram
Answer» The working substance can be air or helium or hydrogen or nitrogen. All processes are reversible. 1. AB is isothermal expansion, at temperature T_H , in which heat Q_H is absorbed from the source and useful work is done by the working substance. 2. BC is isochoric process in which some heat is released by the gas (working substance) to the refrigerator and the gas cools to temperature T_C < T_H . 3. CD is isothermal compression, at temperature T_C , in which heat Q_C is rejected to the coolant (sink). 4. DA is isochoric process in which heat is taken in by the gas and its temperature rises to T_H [Note : Stirling engine operated in reverse direction is used in the field of cryogenics to obtain extremely low – temperatures to liquefy air or other gases.]

3042.

Draw a neat labelled diagram of a Sterling cycle and describe the various processes taking place in a Sterling engine.Sterling cycle P - V diagram

Answer»

The working substance can be air or helium or hydrogen or nitrogen. All processes are reversible.

1. AB is isothermal expansion, at temperature T_H , in which heat Q_H is absorbed from the source and useful work is done by the working substance.

2. BC is isochoric process in which some heat is released by the gas (working substance) to the refrigerator and the gas cools to temperature T_C < T_H .

3. CD is isothermal compression, at temperature T_C , in which heat Q_C is rejected to the coolant (sink).

4. DA is isochoric process in which heat is taken in by the gas and its temperature rises to T_H

[Note : Stirling engine operated in reverse direction is used in the field of cryogenics to obtain extremely low – temperatures to liquefy air or other gases.]

Discussion

3043.	The coefficient of performance of a Carnot refrigerator is 4. If the temperature of the hot reservoir is 47 °C, find the temperature of the cold reservoir.
Answer» Data : K = 4, T_H = 47°C = (273 + 47) K = 320 K = \(\frac{T_C}{T_H-T_C}\) = \(\frac{(4)((320)}{1+4}\)K = (0.8)(320)K = 256K = (256 – 273)°C = – 17°C This is the temperature of the cold reservoir. [Note : A hot-air type engine consisting of two cylinders, was developed by Robert Stirling (1790 -1878), a Scottish engineer and clergyman. He developed the concept in 1816 and obtained a patent for his design in 1827. Some engines were made in 1844. He also used helium and hydrogen in some engines developed thereafter. Stirling engines are used in submarines and spacecrafts.]

3043.

The coefficient of performance of a Carnot refrigerator is 4. If the temperature of the hot reservoir is 47 °C, find the temperature of the cold reservoir.

Answer»

Data : K = 4, T_H = 47°C = (273 + 47) K = 320

K = \(\frac{T_C}{T_H-T_C}\) = \(\frac{(4)((320)}{1+4}\)K = (0.8)(320)K

= 256K = (256 – 273)°C = – 17°C

This is the temperature of the cold reservoir.

[Note : A hot-air type engine consisting of two cylinders, was developed by Robert Stirling (1790 -1878), a Scottish engineer and clergyman. He developed the concept in 1816 and obtained a patent for his design in 1827. Some engines were made in 1844. He also used helium and hydrogen in some engines developed thereafter. Stirling engines are used in submarines and spacecrafts.]

Discussion

3044.	The coefficient of performance of a Carnot refrigerator working between `30^(@)C` and `0^(@)C` isA. 10B. 1C. 9D. 0
Answer» Correct Answer - C

Discussion

3045.	The coefficient of performance of a Carnot refrigerator is 4. If TC = 250 K, then TH = (A) 625 K (B) 310 K (C) 312.5 K (D) 320 K.
Answer» Correct option is (C) 312.5 K

Discussion

3046.

Differentiate between Gas and vapour.

Answer»

Gas	vapour
1. A gas below the critical temperature for the substance is a vapour.	A vapour abve the critical temperature is a gas.
2. A gas cannot be liquefied by the application of pressure alone, however large the pressure may be.	A vapour can be liquefied under pressure alone.
3. Obeys the gas laws.	Does not Obeys the gas laws.

Discussion

3047.	`DeltaU+DeltaW=0`is valid forA. Adiabatic processB. Isochoric processC. Isobaric processD. Isothermal process
Answer» Correct Answer - A

Discussion

3048.	The process in which no heat enters or leaves the system is termed asA. IsochoricB. IsobaricC. IsothermalD. Adiabatic
Answer» Correct Answer - D

Discussion

3049.	A heat insulating cylinder with a movable piston contins 5 moles of hydrogen at standard temperature and prssure if the gas is cmpressed to quarter of its original volume then the pressure of the gas is increased by `(gamma=1.4)`A. `(2)^(1.4)`B. `(3)^(1.4)`C. `(4)^(1.4)`D. `(5)^(1.4)`
Answer» Correct Answer - C As the cylinder is a heat insulating i.e no heat is allowed to be exchanged .Hence the process is adiabatic `P_(1)V_(1)^(gamma)=P_(2)V_(2)^(gamma)` `(P_(2))/(P_(1))=(V_(1))/(V_(2))^(gamma)` Here `V_(1)=V,V_(2)=(V)/(4),P_(1)=P` and `gamma=1.4` `(P_(2))/(P)=((V)/(V//4))^(1.4)=(4)^(1.4) rArr P_(2)=(4)^(1.4)P`

Discussion

3050.	In a Carnot engine when `T_(2) = 0^(@)C` and `T_(1) = 200^(@)C` its efficiency is `eta_(1)` and when `T_(1) = 0^(@)C` and `T_(2) = -200^(@)C`. Its efficiency is `eta_(2)`, then what is `eta_(1)//eta_(2)`?A. 0.577B. 0.733C. 0.638D. Can not be calculated
Answer» Correct Answer - A

Discussion

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