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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.
| 2851. |
Consider the reacton `C(s) + (1)/(2) O_(2)(g) rarr CO(g) + 200 kJ` The signs of `Delta S and Delta G` respectively areA. `+, -, -`B. `-, +, +`C. `-, -, -`D. `-, +, -` |
| Answer» Correct Answer - D | |
| 2852. |
A system changes from the state `(P_(1),V_(1))` to `(P_(2)V_(2))` as shwon in the diagram. The workdone by the system is |
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Answer» Correct Answer - `12 xx 10^(5)` J |
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| 2853. |
A monotomic ideal gas of two metal is taken through a cyclic process straining from A as shown `V_(B)//V_(A)=2 and V_(D)//V_(A)=4` Temperature `T_(A) is 27^(@)C` The work done during the process `C to D` is A. 900R (absorbed)B. 900R(released)C. 1200R(absorbed)D. 1200R(released) |
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Answer» Correct Answer - B,D |
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| 2854. |
If `C_(p) and C_(v)` denote the specific heats (per unit mass of an ideal gas of molecular weight `M`), then where `R` is the molar gas constant.A. `C_(p)-C_(v)= R//M^(2)`B. `C_(p)-C_(v)=R`C. `C_(p)-C_(v)=RM`D. `C_(p)-C_(v)= MR` |
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Answer» Correct Answer - C We know that molor specific heat of a gas is molecular weight times specific heat per unit mass of a gas. If `C_(v) and C_(p)` are the molar specific heats of the ideal gas at constant volume and constant pressure respectively, then `C_(p)= M c_(p)` and `C_(v)= M c_(v)` But `C_(p)-C_(v)= R` `:. M c_(p)-M c_(v)=R or c_(p)-c_(v)=R/M` |
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| 2855. |
Calculate the enthalpy of formation of acetic acid if the enthalpy of combustion to `CO_(2)(g)` and `H_(2)O(l)` is `- 867.0 kJ mol^(-1)` and enthalpies of formation of `CO_(2)(g)` and `H_(2)O(l) ` are respectively -393.5 and -285.9 kJ `mol^(-1)` |
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Answer» Correct Answer - `-491.8 kJ mol^(-1)` Aim `: 2C(s) + 2H_(2) (g) + O_(2)(g) rarrCH_(3)COOH,DeltaH = ?` |
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| 2856. |
A cyclinder contains `0.5` mole of an ideal gas at `310K`. As the gas expands isothermally from an initial volume `0.31m^(3)` to a final volume of `0.45m^(3)`, find the amount of heat that must be added to the gas in order to maintain a constant temperature. |
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Answer» Here, `n= 0.5 mol e , T= 310K`, `V_(1)= 0.31m^(3), V_(2)=0.45m^(3), dQ=?` As the process is isothermal, `dU=0` `dW= nRT log_(e)((V_(2))/(V_(1)))` `=0.5xx8.31xx310log_(e)((0.45)/(0.31))` `dW= 0.5xx8.31xx310xx2.303log_(10)((0.45)/(0.31))` `dW= 480J` `dQ=dU+dW=0+480= 480J` `=(480)/(4.2)cal. = 114cal.` |
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| 2857. |
In Fig., a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. the lower compartment of the container is filled with 2 moles of an ideal monoatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. the heat capacities per mole of an ideal monoatomic gas are `C_(upsilon) = (3)/(2) R and C_(P) = (5)/(2) R`, and those for an ideal diatomic gas are `C_(upsilone) = (5)/(2) R and C_(P) = (7)/(2) R.` Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. the total work done by the gases till the time they achieve equilibrium will be A. `250 R`B. `200 R`C. `100 R`D. `-100 R` |
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Answer» Correct Answer - D Let final temperature of both comparments is `T` . Heat given by lower compartment `Q=nC_(p)DeltaT=2xx(5)/(2)Rxx(700-T)` ….(i) Heat obtained by upper compartment `Q=nC_(p)DeltaT=2xx(7)/(2)Rxx(T-400)` By equatin (i) and (ii) `5(700-T)-7(T-400)` `3000-5T=7T-2800` `6300-12T` `T=525 K` Work done by lower gas `W_(L)=PDeltaV=nRDeltaT=-350 R` Work done by upper gas `W_(u)=PDeltaV=nRDeltaT=+250 R` Net work done `W_(L)+W_(U)=-350+250=-100R` |
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| 2858. |
The mixing of gases is generally accompanied byA. Decrease in entropyB. Decrease in free energyC. Change in heat contentD. Increase in free energy |
| Answer» Correct Answer - B | |
| 2859. |
Mixing of non-reacting gases is generally accompanied byA. Decreases in entropyB. Increases in entropyC. Change in enthalpyD. Change in free energy |
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Answer» Correct Answer - B Mixing of non- reacting gases increases randomness and so increase entropy. |
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| 2860. |
Which of the following P-V curve best represents. As isothermal process?A. B. C. D. |
| Answer» Correct Answer - C | |
| 2861. |
Work done in given cyclic process is A. `P_(0)V_(0)`B. `3P_(0)V_(0)`C. `6P_(0)V_(0)`D. `5P_(0)V_(0)`. |
| Answer» Correct Answer - C | |
| 2862. |
In a carnot engine, for `eta=1`, which of the following is true? (Symbols have their usual meaning)A. `T_(1)=T_(2)`B. `T_(1)=0`C. `T_(2)=0`D. `eta` is independent of `T_(1) and T_(2)`. |
| Answer» Correct Answer - C | |
| 2863. |
Assetion : No engine can have efficiencyt greater than that of the carnot engine Reason : The efficiencyt of a cornot engine is given by `eta =1-(T_(2))/(T_(1))`A. If both assetion and reason are true and reaasons is the correct expanation of assetionB. If both assetion and reason are tur but reason is not the correct explanation of assetionC. If assertion is true but reason is falseD. If both assertion and reason are false. |
| Answer» Correct Answer - B | |
| 2864. |
An ice cube of mass `M_(0)` is given a velocity `v_(0)` on a round horizontal surface with coefficient of friction `mu`. The block is at its melting point and latent heat of fusion of ice is `L` . The block receive heat only due to the friction forces and all work is converted into heat. Find the mass of the remaining ice block after time t. A. `m=m_(0)e^(-(2mug)/(L)(V_(0)t+(1)/(2)mu"gt"^(2)))`B. `m=m_(0)e^(-(2mu)/(L)(V_(0)t-(1)/(2)mu"gt"^(2)))`C. `m=m_(0)e^(-(3mu)/(L)(V_(0)t-(1)/(2)mu"gt"^(2)))`D. `m=m_(0)e^(-(2mu)/(L)(V_(0)t-mu"gt"^(2)))` |
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Answer» Correct Answer - B `mu mg v =-((dm)/(dt))L` `implies underset(m_(0))overset(m)(int) (dum)/(m)=-(mug)/(L) underset(0)overset(t)(int)(mu"gt")dt` `implies m=m_(0)e^(-(mug)/(L)(v_(0)t-(1)/(2)mu"gt"^(2))` |
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| 2865. |
An ideal gas of molar mass `M` is enclosed in a vessel of volume of `V` whose thin walls are kept at a constant temperature `T`. At a moment `t = 0` a small hole of area `S` is opened, and the gas starts escaping into vacuum. Find the gas concentration `n` as a function of time `t` if at the initial moment `n (0) = n_0`. |
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Answer» We can assume that all molecules, incident on the hole, leak out. Then, `-dN = - d(nV) = (1)/(4) n lt v gt S dt` or `dn = -n (dt)/(4 v//S lt v gt) = -n (dt)/(tau)` Integrating `n = n_0 e^(-t//tau)`. Hence `lt v gt = sqrt((8 RT)/(pi M))`. |
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| 2866. |
A vessel filled with gas is divided into two equal parts `1` and `2` by a thin heat-insulating partition with two holes. One hole has a small diameter, and the other has a very large diameter (in comparsion with the mean free path of molecules). In part `2` the gas is kept at a temperature `eta` times higher than that of part `1` How will the concentration of molecules in part `2` change and how many times after the large hole is closed ? |
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Answer» If the temperature of the compartment `2` is `eta` times more than that of compartment `1`, it must contain `(1)/(eta)` times less number of molecules since pressure must be the same when the big hole is open. If `M` = mass of the gas in `1` than the mass in `1` than the mass of the gas in `2` must be `(M)/(eta)`. So immediately after the big hole is closed. `n_1^0 = (M)/(mV), n_2^0 = (M)/(m V eta)` where `m` = mass of ach molecule and `n_1^0, n_2^0` are concentrations in `1` and `2`. After the big hole is closed the pressures will differ and concentration will become `n_1` and `n_2` where `n_1 + n_2 = (M)/(m V eta) (1 + eta)` On ther other hand `n_1 lt v_1 gt = n_2 lt v_2 gt` i.e. `n_1 = sqrt(eta) n_2` Thus `n_2 (1 + sqrt(eta)) = (m)/(m V eta) (1 + eta) = n_2^0 (1 + eta)` So `n_2 = n_2^0 (1 + eta)/(1 + sqrt(eta))`. |
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| 2867. |
As a result of a certain process the viscosity coefficient of an ideal gas increases `alpha=2.0` times and its diffusion coefficient `beta = 4.0` times. How does the gas pressure change and how many times ? |
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Answer» We know `eta = (1)/(2) lt v gt lamda rho = (1)/(3) lt v gt (1)/(sqrt(2) pi d^2) m alpha sqrt(T)` Thus `eta` changing `alpha` times implies `T` changing `alpha^2` times. On the other hand `D = (1)/(3) lt v gt = (1)/(3) sqrt((8 kT)/(pi m)) (kT)/(sqrt(2) pi d^2 p)` Thus `D` changing `beta` means `(T^(3//2))/(p)` changing `beta` times So `p` must change `(alpha^3)/(beta)` times. |
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| 2868. |
How will a diffusion coefficient `D` and the viscosity coefficient `eta` of an ideal gas change it its volume increases `n` times : (a) isothermally , (b) isobarically ? |
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Answer» `D alpha(sqrt(T))/(n) alpha V sqrt(T), eta alpha sqrt(T)` (a) `D` will increase `n` times `eta` will remain constant if `T` is constant (b) `D alpha (T^(3//2))/(p) alpha((p V)^(3//2))/(p) = p^(1//2) V^(3//2)` `eta alpha sqrt(p V)` Thus `D` will increase `n^(3//2)` times, `eta` will increase `n^(1//2)` times, if `p` is constant. |
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| 2869. |
Demonstrate that the process in which the work performed by an ideal gas is alphaortional to that corresponding increment of its internal energy is decribed by the equation `p V^n - const`, where `n` is a constant. |
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Answer» According to the problem : `A alpha U` or `dA = aU` (where `a` is alphaortionality constant) or, `pdV = (a v R dT)/(gamma - 1)`…(1) ltrgt From ideal gas law, `pV = v R T`, on differentiating `pdV + Vdp = v RdT`…(2) Thus from (1) and (2) `pdV = (a)/(gamma - 1) (pdV + Vdp)` or, `pdV ((a)/(gamma - 1) - 1) + (a)/(gamma - 1) V dp = 0` or, `pdV(k - 1) + kVdp = 0`(where `k = (a)/(gamma - 1) = "another constant")` or, `pdV (k - 1)/(k) + Vdp = 0` or, `pdVn + Vdp = 0`(where `(k - 1)/(k) = n = ratio`) Diving both the sides by `pV` `n (dV)/(V)+(dp)/(p) = 0` On integrating `n 1n V + 1 n p = 1n C` (where `C` is constant) or, `1 n(pV^n) = 1n C` or, `pV^n = C (const)`. |
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| 2870. |
In a certain polytropic process the volume of argon was increased `alpha= 4.0` times. Simultaneously, the pressure decreased `beta = 8.0` times. Find the molar heat capacity of argon in this process. Assuming the gas to be ideal. |
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Answer» Let the process be polytropic according to the law `pV^n = constant` Thus, `p_f V_f^n = p_i V_i^n` or, `((p_i)/(p_f)) = beta` So, `alpha^n = beta` or `1n beta = n 1n alpha` or `n = (1n beta)/(1 n alpha)` In the polytropic process molar heat capacity is given by `C_n (R(n - gamma))/((n - 1)(gamma - 1)) = (R)/(gamma - 1) -(R)/(n - 1)` =`(R)/(gamma - 1) = (R 1n alpha)/(1 n beta - 1n alpha)`, where `n = (1 nbeta)/(1n alpha)` So, `C_n = (8.314)/(1.66 0-1)-(8.314 1n 4)/(1 n 8 - 1n 4) = -42 J//mol.K`. |
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| 2871. |
An ideal gas consists of rigid diatomic molecules. How will a diffusion coefficient `D` and viscosity coefficient `eta` change and how many times if the gas volume is decreased adiabatically `n = 10` times ? |
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Answer» `D alpha V sqrt(T), eta alpha sqrt(T)` In an adiabatic process `TV^(gamma - 1) = constant`, or `T alpha V^(1 -gamma)` Now `V` is decreased `(1)/(n)` times. Thus `D alpha V^((3 - gamma)/(2)) = ((1)/(n))^((3 - gamma)/(2)) = ((1)/(n))^(4//5)` `eta alpha of V^((1 - gamma)/(2)) = ((1)/(n))^(-1//5) = n^(1//5)` So `D` decrease `n^(4//5)` times and `eta` increase `n^(1//5)` times. |
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| 2872. |
An ideal gas goes through a polytropic process. Find the polytropic exponent `n` if in this process the coefficient (a) of diffusion , (b) of viscosity , ( c) of heat conductivity remains constant. |
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Answer» (a) `D alpha V sqrt(T) alpha sqrt(p V^3)` Thus `D` remains constant in the process `pV^3 = constant` So polytropic index `n = 3` (b) `eta alpha sqrt(T) alpha sqrt(pV)` So `eta` remains constant in the isothermal process `pV = constant, n = 1 here` ( c) Heat conductivity `k = eta C_V` and `C_V` is a constant for the ideal gas Thus `n = 1` here also. |
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| 2873. |
Find the molar heat capacity of an ideak gas in a polytropic process `p V^n = const` if the adiabatic exponent of the gas is equal to `gamma`. At what values of the polytropic constant `n` will the heat capacity of the gas be negative ? |
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Answer» In the polytropic process work done by the gas `A = (vR[T_i - T_f])/(n - 1)` (where `T_i` and `T_f` are initial and final temperature of the gas like in adiabatic process) and `Delta U = (vR)/(gamma - 1)(T_f - T_i)` By the first law of thermodynamics `Q = Delta U + A` =`(vR)/(gamma - 1)(T_f - T_i)+(vR)/(n -1) (T_i - T_f)` =`(T_f - T_i) vR[(1)/(gamma - 1) -(1)/(n -1)] = (vR[n - gamma])/((n - 1)(gamma -1)) Delta T` According to definition of molar heat capacity when number of moles `v = 1` and `Delta T = 1` then `Q` = Molar heat capacity. Here, `C_n = (R(n - gamma))/((n - 1)(gamma - 1)) lt 0 "for" 1 lt n lt gamma`. |
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| 2874. |
If `CH_(3)COOH+OH^(-)toCH_(3)COO^(-)+H_(2)O+q_(1)and H^(+)+OH^(-)to H_(2)O+q_(2)`, then the enthalpy change for the reaction `CH_(3)COOHtoCH_(3)COO^(-)+H^(+)` is "equal to :A. `q_(1)+q_(2)`B. `q_(1)-q_(2)`C. `q_(2)-q_(1)`D. `-q_(1)-q_(2)` |
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Answer» Correct Answer - c |
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| 2875. |
A certain mass of gas at 273 K is expanded to 81 times its volume under adiabatic condition. If `lambda=1.25` for gas , then its final temperature isA. `-235^(@)C`B. `-182^(@)C`C. `-91^(@)C`D. `-0^(@)C` |
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Answer» Correct Answer - B |
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| 2876. |
which reaction proceeds with the greates increases in entropyA. `H_(2)(g) +O_(2)(g) to H_(2)O_(2)(l) `B. `Br_(2)(l) +F_(2)(g) to 2BrF(g)`C. `Cu^(2+)(aq) +Zn(s) to Cu(s) + Zn^(2+)(aq)`D. `4NH_(3)(g) + 7O_(2)(g) to 4NO_(2)(g) + 6H_(2)O(g) ` |
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Answer» Correct Answer - B |
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| 2877. |
One mole of an ideal gas with the adiabatic exponent `gamma` goes through a polytropic process as a result of which the absolute temperature of the gas increases `tau-fold`. The polytropic constant equal `n`. Find the entropy increment of the gas in this process. |
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Answer» For the polytropic process with index `n` `p V^n = constant` Along this process (Sec 2.122) `C = R((1)/(gamma -1) -(1)/(n - 1)) = (n - gamma)/((gamma -1)(n - 1)).R` So `Delta S = int _(T_0)^(tau_(T_0)) C (dT)/(T) = (n - gamma)/((gamma - 1)(n - 1)) R 1n tau`. |
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| 2878. |
The expansion process of `v = 2.0` moles of argon proceeds so that the gas pressure increases in direct alphaortion to its volume. Find the entropy increment of the gas in this process provided its volume increases `alpha = 2.0`times. |
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Answer» The process in question may be written as `(p)/(p_0) = alpha (V)/(V_0)` where `alpha` is a constant and `p_0, V_0` are some reference values. For this process (see 2.127) the specific heat is `C = C_V + (1)/(2) R = R ((1)/(gamma - 1) + (1)/(2)) = (1)/(2) R (gamma + 1)/(gamma - 1)` Along the line volume increases `alpha` times then so does so does the pressure. The temperature must then increase `alpha^2` times. Thus `Delta S = int _(T_0)^(alpha^2 T_0) vC (dT)/(T) = (vR)/(2) (gamma + 1)/(gamma - 1) 1n alpha^2 = vR (gamma + 1)/(gamma - 1) 1n alpha` if `v = 2, gamma = (5)/(3), alpha = 2, Delta S = 46.1 "Joule"//.^@K`. |
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| 2879. |
A refrigerator works between `4^(@)C` and `30^(@)C`. It is required to remove `600 calories` of heat every second in order to keep the temperature of the refrigerator space constant.The power required is (Take `1calorie= 4.2 J`)A. `23.65W`B. `236.5W`C. `2365W`D. `2.365W` |
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Answer» Correct Answer - B Here,`T_(2)= 4^(@)C= 4+273= 277K` `T_(1)= 30^(@)C=30+273= 303K` `Q_(2)= 600cal//s= 600xx4.2J//s` `P=?` As `COP=(Q_(2))/W= (T_(2))/(T_(1)-T_(2))` `:. W=(Q_(2)(T_(1)_T_(2)))/(T_(2))= (600xx4.2(303-277))/(277)` `W= 236.5 J//s= 236.5 "Watt"`. |
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| 2880. |
The entropy of a gas increases on its expansion . Why? |
| Answer» Correct Answer - Because larger space creates more disorder. | |
| 2881. |
the heat of atomization of methane and ethane are 360 KJ/mol and 620 KJ/mol , respectively . The longest Wavelength of light capable of breaking . The c-c bond is : (Avogadro number`=6.02xx10^(23),h=6.62xx10^(-34)Js)`A. `2.48 xx 10^(-3)` nmB. `1.49 xx 10^(3)` nmC. `2.49 xx 10^(4)` nmD. `2.48 xx 10^(4)` nm |
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Answer» Correct Answer - B `CH_(4)(g)rarr C(g) + 4H(g)` `implies " " 4xx E_(C-H) = 360 KJ//Mol." " implies E_(C-H)=90KJ//Mol.` and `C_(2)H_(6)(g) rarr 2C(g) + 6H(g)` `E_(C-C)+ 6 xx90 = 920 " " implies" " E_(C-C)=80 KJ//mol` ` implies N_(A) xx (hc)/(lambda) = 80 xx 1000 J` `lambda =(6.02 xx 10^(23) xx 6.62xx 10^(-34) xx 3 xx 10^(8))/(80000) = 14.9 xx 10^(-7)m = 1.49 xx 10^(-6)m =1.49 xx 10^(3)` nm |
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| 2882. |
the heat of atomization of methane and ethane are 360 KJ/mol and 620 KJ/mol , respectively . The longest Wavelength of light capable of breaking . The c-c bond is : (Avogadro number`=6.02xx10^(23),h=6.62xx10^(-34)Js)`A. `1.49xx10^(3)cm`B. `2.48xx10^(4)nm`C. `2.48xx10^(3)nm`D. `1.49xx10^(4)nm` |
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Answer» Correct Answer - a |
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| 2883. |
Which change(s) is(are) accompanied by an increase in entropy of the system? (P) Conversion of `O_(2)(g) "to" O_(3)(g)` (Q) Freezing of water (R ) Sublimation of iodineA. P onlyB. R ONLYC. P AND Q ONLYD. Q and R only |
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Answer» Correct Answer - B |
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| 2884. |
Which of the following is not a state function? (a) S (b) H (c) G (d) q |
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Answer» Answer: (d) q |
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| 2885. |
Which of the following will have highest ∆Hvap Value? (a) Acetone (b) Ethanol (c) Carbon tetrachloride (d) Chloroform |
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Answer» Answer: (b) Ethanol |
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| 2886. |
Which of the following is a state function? (a) q (b) ∆q (c) w (d) ∆S |
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Answer» Answer: (d) ∆S |
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| 2887. |
In which of the following entropy increases? (a) Condensation of water vapour (b) Liquid freezes to solid (c) Sublimation (d) Gas freezes to a solid |
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Answer» (c) Sublimation |
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| 2888. |
Two moles of a triatomic linear gas (neglect vibration degree of freedom) are taken through a reversible process ideal starting from A as shown in figure. The volume ratio `(V_(B))/(V_(A))=4`. If the temperature at A is `-73^(@)C`, then : Work done by the gas in AB process is : `6.16` kJ `308.3` kJ `9.97` kJ 0 J (ii) Total enthalpy change in both steps is :A. 3000 RB. 4200 RC. 2100 RD. 0 |
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Answer» Correct Answer - B::C (i) (c) `w=-P.DeltaV=-nRDeltaT=-2xx8.314xx600` `=-9.97kJ` (ii) (b) `DeltaH_("total")=DeltaH_(AB)+DeltaH_(BC)=nC_(p,m)DeltaT+0` `=2xx(7)/(2)xxRxx(800-200)` = 4200 R |
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| 2889. |
I st law can explainA. Spontaneity of processB. Perpetual machine of 2 nd kindC. Perpetual machine of 3 rd kindD. The total energy of an isolated system remains constant |
| Answer» Correct Answer - D | |
| 2890. |
Which of the following is an extensive property? (a) Volume, (b) Surface tension, (c) Viscosity, (d) Density |
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Answer» (a) Volume is an extensive property. |
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| 2891. |
For the reaction, N2(g) + 3H2(g) → 2NH3(g) predicts whether the work is done on the system or by the system. |
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Answer» Volume is decreasing therefore, work is done on the system. |
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| 2892. |
What is the limitation of first law of thermodynamics? |
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Answer» It cannot tell us the direction of the process. |
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| 2893. |
According to 1 st law of ThermodynamicsA. Energy can be created but not destroyedB. Energy cannot be created but can be destroyedC. energy can be created and destroyedD. Energy can not be created nor destroyed |
| Answer» Correct Answer - D | |
| 2894. |
Acoording to 1 st law of ThermodynamicsA. The energy of system is constantB. The energy of universe is constantC. The energy of surroundings is constantD. The energy of system and surroundings are not constant |
| Answer» Correct Answer - B | |
| 2895. |
Heat energy absorbed by a system in going through a cyclic process shown in Fig. A. `10pij`B. `10pij`C. `10pij`D. `10^(-3)pij` |
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Answer» Correct Answer - C |
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| 2896. |
Which of the process described below are irrevesible?A. The increase in temperature of an iron rod by hammering it.B. A gas in small container at a temperature `T_(1)` is brought in contact with a big reservoir at a higher temperature of the gas.C. A quasi- static isothermal expansion of an ideal gas in cyclinder fitted with a frictionless piston.D. An ideal gas is enclosed in a piston cyclinder arrangement with adiabatic walls. A weight `W` is added to the piston, resulting in compression of gas. |
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Answer» Correct Answer - A::B::D (a) The increase in temperature of an iron rod by hammering it is the irrevesible process. (b) 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 gas is also an irreversible process. (d) An ideal gas is enclosed in a piston cyclinder arrangement with adiabatic walls. A weight `W` is added to piston resulting in compression of the gas is also an irreversible process. In all the cases above, the original state of the system cannot be regained. |
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| 2897. |
An exothermic reaction is one in which the reacting substances :A. Have same energy as productsB. Have less energy than the productsC. Have more energy than the productsD. Are at higher temperature than the products |
| Answer» Correct Answer - C | |
| 2898. |
In endothermic reactions the reactants :A. Have more energy than productsB. Have as much energy as the productsC. Are at lower temperature than productsD. Have less enregy than the products |
| Answer» Correct Answer - D | |
| 2899. |
The `DeltaH_(f)^(@)` of MgO is `-602KJxxmol^(-1)` .when 20.15g MgO is decomposed at constant pressure according to the equation below , how much heat will be transferred? `2MgO(s)to 2MG(s)+O_(2)(g)`A. `1.20xx10^(3)KJ"of heat is released "`B. `6.02xx10^(2)KJ"of heat is absorbed "`C. `6.02xx10^(2)KJ"of heat is released "`D. `3.01xx10^(2)KJ"of heat is absorbed "` |
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Answer» Correct Answer - d |
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| 2900. |
What mass of ice at `0.0^(@)` C must be added to 100g `H_(2)O` at `25.0^(@)C` to cool It to `0.0^(@)`C? The heat of fusion of ice is `334Jg^(-1)`A. `1.25g`B. `7.49g`C. `31.3g`D. 100g |
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Answer» Correct Answer - c |
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