54 + Interview Questions in THERMODYNAMIC IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Calculate the enthalpy change when `50 mL` of `0.01 M Ca(OH)_(2)` reacts with `25mL` of `0.01 M HCI`. Given that `DeltaH^(Theta)` neutralisaiton of strong acid and string base is `140 kcal mol^(-1)`A. `14 cal`B. `35 cal`C. `10 cal`D. `7.5 cal`
Answer» Correct Answer - 2 `m.eq.` of base `=50xx0.01xx2=1` `m.eq.` of acid `=25xx0.01" "0.25` `:. Me.eq.` of base reacted `=0.25` `:.` energy released `=(140)/(1000)xx0.25Kcal=(140xx0.25xx1000)/(1000)cal=35cal`.

Discussion

2.	If an average person jogs, hse produces 14.5 × 103 cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 103 cal for evaparation) is(a) 0.25 kg(b) 2.25 kg(c) 0.05 kg(d) 0.20 kg
Answer» The amount of sweat evaporated per minute (a) 0.25 kg

2.

If an average person jogs, hse produces 14.5 × 103 cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 103 cal for evaparation) is(a) 0.25 kg(b) 2.25 kg(c) 0.05 kg(d) 0.20 kg

Answer»

The amount of sweat evaporated per minute

(a) 0.25 kg

Discussion

3.	For the reaction at `25^(@)C, C_(2)O_(4)(l) rarr 2XO_(2)(g)` `Delta H =2.1 kcal` and `DeltaS=20 cal K^(-1)` . The reaction would be `:`A. spontaneousB. non`-` spontaneousC. at equilibriumD. unpredictable
Answer» Correct Answer - 1 `DeltaG=DeltaH-TDeltaS` `=2.1xx10^(3)-20xx298lt0`

Discussion

4.	A reaction has `DeltaH=-33KJ` and `DeltaS=-58(J)/(K)` . This reaction would be:A. spontaneous at all temperatureB. non`-` spontaneous at all temperaturesC. spontaneous above a certain temperature onlyD. spontaneous below a certain temperature only
Answer» Correct Answer - 4 `{:(DeltaG,=,(DeltaH),-,T(DeltaS)),(,,darr,,darr),(,,-ve,,-ve):}` since both are `-ve, ` the reaction would have a `-ve DeltaG` below a temperature of `(33000)/(58)K(=569K)`

Discussion

5.	During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is
Answer» The correct answer is (iii)

Discussion

6.	Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options (i) to (iv) is correct?(a) C(g) + 4H(g) → CH4 (g); ΔrH = x kJ mol–1(b) C (graphite,s) + 2H2 (g) → CH4 (g); ΔrH = y kJ mol–1(i) x = y(ii) x = 2y(iii) x > y(iv) x < y
Answer» (iii) Justification : Same bonds are formed in reaction (a) and (b) but bonds between reactant molecules are broken only in reaction (b).

Discussion

7.	The entropy change can be calculated by using the expression ΔS = qrev/T . When water freezes in a glass beaker, choose the correct statement amongst the following : (i) ΔS (system) decreases but ΔS (surroundings) remains the same. (ii) ΔS (system) increases but ΔS (surroundings) decreases. (iii) ΔS (system) decreases but ΔS (surroundings) increases. (iv) ΔS (system) decreases and ΔS (surroundings) also decreases.
Answer» (iii) Justification : Freezing is exothermic process. The heat released increases the entropy of surrounding.

Discussion

8.	The enthalpy of combustion of graphite is 393.4 kJ. Calculate(a) The amount of graphite needed to produce 196.7 KJ of heat.(b) The number of moles of CO2 formed when 196.7 KJ of heat is produced.(c) The volume of oxygen required at S.T.P. to from 24.0 g of graphite in this process.
Answer» We are given (i) C(graphite) + O₂(g) → CO₂(g); ΔH=-393.4KJ (a) From the above equation, we know that 393.4 KJ of heat is produced by 12 g of graphite. 196.7 kj of heat is produced by 12/393.4 x 196.7= 6 g of graphite (b) From equation (i) we can say that Production of 393.4 KJ of heat is accompanied by the formation of 1 mole of CO₂. Production of 196.7 KJ of heat is accompanied by the formation of 0.5 mole of CO₂. (C) Volume of oxygen required at S.T.P to burn 12 g of graphite = 22.4 litres. Volume of oxygen required at S.T.P to burn 24 of graphite = 22.4 X 2 = 44.8 litres.

8.

The enthalpy of combustion of graphite is 393.4 kJ. Calculate(a) The amount of graphite needed to produce 196.7 KJ of heat.(b) The number of moles of CO2 formed when 196.7 KJ of heat is produced.(c) The volume of oxygen required at S.T.P. to from 24.0 g of graphite in this process.

Answer»

We are given

(i) C(graphite) + O₂(g) → CO₂(g); ΔH=-393.4KJ

(a) From the above equation, we know that 393.4 KJ of heat is produced by 12 g of graphite.

196.7 kj of heat is produced by

12/393.4 x 196.7= 6 g of graphite

(b) From equation (i) we can say that

Production of 393.4 KJ of heat is accompanied by the formation of 1 mole of CO₂.

Production of 196.7 KJ of heat is accompanied by the formation of 0.5 mole of CO₂.

(C) Volume of oxygen required at S.T.P to burn 12 g of graphite = 22.4 litres.

Volume of oxygen required at S.T.P to burn 24 of graphite = 22.4 X 2 = 44.8 litres.

Discussion

9.	The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction. H2(g) + Br2(g) → 2HBr(g) Given that Bond energy of H2, Br2 and HBr is 435 kJ mol–1, 192 kJ mol–1 and 368 kJ mol–1 respectively.
Answer» Δ_rH^Θ = Bond energy of H₂ + Bond energy of Br₂ – 2 × Bond energy of HBr = 435 + 192 – (2 × 368) kJ mol^–1 ⇒ Δ_rH^Θ = –109 kJ mol^–1

9.

The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction. H2(g) + Br2(g) → 2HBr(g) Given that Bond energy of H2, Br2 and HBr is 435 kJ mol–1, 192 kJ mol–1 and 368 kJ mol–1 respectively.

Answer»

Δ_rH^Θ = Bond energy of H₂ + Bond energy of Br₂ – 2 × Bond energy of HBr

= 435 + 192 – (2 × 368) kJ mol^–1

⇒ Δ_rH^Θ = –109 kJ mol^–1

Discussion

10.	For a closed container containing `100` mol of an ideal gas fitted with movable, frictionless, weightless piston operating such that pressure of gas remain constant at `8.21` atm, which graph repsents correct variation of log `V v//s` log `T` where `V` is in litre and `T` is in Kelvin ? `(R = 0.0821 (atmL)/(molK))`A. B. C. D.
Answer» Correct Answer - 1 `V=(nRT)/(P)=(100xx0.0821xxT)/(8.21)=T` `:." "logV=logT`

Discussion

11.	The enthalpy of neutralization of `HNO_(3)` by `NaOH=-13680 cal//eqt.` When one equivalent of `NaOH` is added to a dilute solution containing one equivalent of `HNO_(3)` and one equivalent of a certain monoprotic weak acid `13960` cals are evolved. Assume that the base is distributed between `HNO_(3)` and the weak acid in the ratio `3:1` and the weak acid is practically nonionized. Calculate the enthalpy of ionization of the weak acid.A. `-1120` calsB. `-2110` calsC. `1210` calsD. `+1210 cals`
Answer» Correct Answer - 3 `13960=-13680xx(3)/(4)-13680xx(1)/(4)+(1)/(4)x` where x is enthalpy of dissociation/ionisation of weak acid. so `x=1120` calories.

Discussion

12.	What is the final temperature of `0.10` mole monoatomic ideal gas that performs `75 cal` of work adiabatically.if the initial temperature is `227 ^(@)C` ( use `R=2 cal //K-mol)`A. `250K`B. `300K`C. `350K`D. `750K`
Answer» Correct Answer - 1 `DeltaU=w` `rArr" "-75=0.1xx(3)/(2)xx[T_(2)-500]` `T_(2)=250K`

Discussion

13.	At 298 K. Kp for the reaction N2O4 (g) ⇄ 2NO2 (g) is 0.98. Predict whether the reaction is spontaneous or not.
Answer» Δ_rG° = -RT ln K_p = -RT ln (0.98) Since In (0.98) is negative .’. Δ G° is positive => the reaction is non spontaneous

13.

At 298 K. Kp for the reaction N2O4 (g) ⇄ 2NO2 (g) is 0.98. Predict whether the reaction is spontaneous or not.

Answer»

Δ_rG° = -RT ln K_p

= -RT ln (0.98)

Since In (0.98) is negative

.’. Δ G° is positive

=> the reaction is non spontaneous

Discussion

14.	A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in Fig. 6.1. What will be the value of ΔH for the cycle as a whole?
Answer» For a cyclic process, ΔH = 0

Discussion

15.	18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol–1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?
Answer» Enthalpy of a reaction is the energy change per mole for the process. 18 g of H₂O = 1 mole (ΔH_vap = 40.79 kJ moE¹ ) Enthalpy change for vapourising 2 moles of H₂O = 2 x 40.79 = 81.58 kJ ΔH°_vap = 40.79 kJ mol^-1

15.

18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol–1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?

Answer»

Enthalpy of a reaction is the energy change per mole for the process.

18 g of H₂O = 1 mole (ΔH_vap = 40.79 kJ moE¹ )

Enthalpy change for vapourising 2 moles of H₂O = 2 x 40.79 = 81.58 kJ ΔH°_vap = 40.79 kJ mol^-1

Discussion

16.	Which quantity out of ΔrG and ΔrGΘ will be zero at equilibrium?
Answer» Δ_rG will always be zero. Δ_rG^Θ is zero for K = 1 because ΔG^Θ = – RT lnK, ΔG^Θ will be non zero for other values of K.

16.

Which quantity out of ΔrG and ΔrGΘ will be zero at equilibrium?

Answer»

Δ_rG will always be zero.

Δ_rG^Θ is zero for K = 1 because ΔG^Θ = – RT lnK, ΔG^Θ will be non zero for other values of K.

Discussion

17.	Identify the state functions and path functions out of the following :enthalpy, entropy, heat, temperature, work, free energy.
Answer» State Functions : Enthalpy, Entropy, Temperature, Free energy Path Functions : Heat, Work

Discussion

18.	One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?
Answer» Water has higher enthalpy of vapourisation.(ΔH_r)_water > (ΔH_r)_acetone

Discussion

19.	The molar enthalpy of vapourisation of acetone is less than that of water. Why?
Answer» Because of strong hydrogen bonding in water, its enthalpy of vapourisation is more.

Discussion

20.	Thermodynamics mainly deals with(i) interrelation of various forms of energy and their transformation from one form to another.(ii) energy changes in the processes which depend only on initial and final states of the microscopic systems containing a few molecules.(iii) how and at what rate these energy transformations are carried out.(iv) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.
Answer» (i) interrelation of various forms of energy and their transformation from one form to another. (iv) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.

20.

Thermodynamics mainly deals with(i) interrelation of various forms of energy and their transformation from one form to another.(ii) energy changes in the processes which depend only on initial and final states of the microscopic systems containing a few molecules.(iii) how and at what rate these energy transformations are carried out.(iv) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.

Answer»

(i) interrelation of various forms of energy and their transformation from one form to another.

(iv) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.

Discussion

21.	Match the following :(i) Entropy of vapourisation(a) decreases(ii) K for spontaneous process(b) is always positive(iii) Crystalline solid state(c) lowest entropy(iv) ΔU in adiabatic expansion of ideal gas(d) ΔHvap/Tb
Answer» (i) → (b), (d) (ii) → (b) (iii) → (c) (iv) → (a)

Discussion

22.	Consider the reaction at `300K` `C_(6)H_(6)(l)+(15)/(2)O_(2)(g)rarr 6CO_(2)(g)+3H_(2)O(l),DeltaH=-3271`kJ What is `DeltaU` for the combustion of `1.5` mole of benzene at `27^(@)C` ?A. `-3267.25kJ`B. `-4900.88kJ`C. `-4906.5kJ`D. `-3274.75kJ`
Answer» Correct Answer - 2 For 1 mole of combustion of benzene `Deltan=-1.5` `DeltaH^(@)=DeltaU+Deltan_(9)RT` `rArr-3271=DeltaU-(1.5xx8.314xx300)/(1000)` `rArr" "DeltaU=-3267.25xx1.5=-4900.88kJ`

Discussion

23.	In an exothermic reaction, heat is evolved, and system loses heat to the surrounding. For such system(i) qp will be negative(ii) ΔrH will be negative(iii) qp will be positive(iv) ΔrH will be positive
Answer» (i) q_p will be negative (ii) Δ_rH will be negative

23.

In an exothermic reaction, heat is evolved, and system loses heat to the surrounding. For such system(i) qp will be negative(ii) ΔrH will be negative(iii) qp will be positive(iv) ΔrH will be positive

Answer»

(i) q_p will be negative

(ii) Δ_rH will be negative

Discussion

24.	In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.Assertion (A): Combustion of all organic compounds is an exothermic reaction.Reason (R) : The enthalpies of all elements in their standard state are zero.(i) Both A and R are true and R is the correct explanation of A.(ii) Both A and R are true but R is not the correct explanation of A.(iii) A is true but R is false.(iv) A is false but R is true.
Answer» (ii) Both A and R are true but R is not the correct explanation of A.

Discussion

25.	Explain whya) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1+ T2)/2.b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.c) Air pressure in a car tyre increases during driving.d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Answer» a) When two bodies at different temperatures T₁ and T₂ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T₁ + T₂)/2 only when the thermal capacities of both the bodies are equal. b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot. c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase. d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

25.

Explain whya) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1+ T2)/2.b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.c) Air pressure in a car tyre increases during driving.d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Answer»

a) When two bodies at different temperatures T₁ and T₂ are brought in thermal

contact, heat flows from the body at the higher temperature to the body at the

lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T₁ + T₂)/2 only when the thermal capacities of both the bodies are equal.

b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.

c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.

d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

Discussion

26.	A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig. 12.11. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R)AB : constant volumeBC : constant pressureCD : adiabaticDA : constant pressure
Answer» Q_AB = U_AB = (3/2)R(T_B - T_A) = (3/2) Q_Bc = U_BC + W_BC = (3/2) P_B(V_C - V_B) + P_B(V_C - V_B) = (5/2) P_B(V_C - V_A) Q_CA = 0 Q_DA = (5/2) P_A (V_A - V_D)

26.

A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig. 12.11. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R)AB : constant volumeBC : constant pressureCD : adiabaticDA : constant pressure

Answer»

Q_AB = U_AB = (3/2)R(T_B - T_A) = (3/2)

Q_Bc = U_BC + W_BC

= (3/2) P_B(V_C - V_B) + P_B(V_C - V_B)

= (5/2) P_B(V_C - V_A)

Q_CA = 0

Q_DA = (5/2) P_A (V_A - V_D)

Discussion

27.	If a air is a cylinder is suddenly compressed by a piston. What happens to the pressure of air?
Answer» Since the sudden compression causes heating and rise in temperature and if the piston is maintained at same Position then the pressure falls as temperature decreases.

Discussion

28.	Air pressure in a car tyre increases during driving. Explain.
Answer» During driving, temperature of the gas increases while its volume remains constant. So according to Charle’s law, at constant V, P α T. Therefore, pressure of gas increases.

28.

Air pressure in a car tyre increases during driving. Explain.

Answer»

During driving, temperature of the gas increases while its volume remains constant.

So according to Charle’s law, at constant V, P α T.

Therefore, pressure of gas increases.

Discussion

29.	What is the ratio of slopes of P-V graphs of adiabatic and isothermal process?
Answer» The slope of P-V graph is dP/dV For an isothermal process, (PV = constant) So dP/dV = P/V → (1) For an adiabatic process ( PV^Y = constant) dP/dV = YP/V → (2) Divide 2) by 1) So, the ratio of adiabatic slope to isothermal slope is Y.

29.

What is the ratio of slopes of P-V graphs of adiabatic and isothermal process?

Answer»

The slope of P-V graph is dP/dV

For an isothermal process, (PV = constant)

So dP/dV = P/V → (1)

For an adiabatic process ( PV^Y = constant)

dP/dV = YP/V → (2)

Divide 2) by 1)

So, the ratio of adiabatic slope to isothermal slope is Y.

Discussion

30.	In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.Assertion (A) : A liquid crystallises into a solid and is accompanied by decrease in entropy.Reason (R) : In crystals, molecules organise in an ordered manner.(i) Both A and R are true and R is the correct explanation of A.(ii) Both A and R are true but R is not the correct explanation of A.(iii) A is true but R is false.(iv) A is false but R is true.
Answer» (i) Both A and R are true and R is the correct explanation of A.

Discussion

31.	Explain the Heat of formation acetylene is +54 kcal.
Answer» When 1 mole of gaseous acetalylene is formed from its elements, carbon (s) and hydrogen (g), 54 kcal of heat is absorbed.

Discussion

32.	No real engine can have an efficiency greater that of a carnot engine working between the same low temperatures. Why ?
Answer» In carnot engine. (i) There is absolutely no friction between the wall of cylinder and piston. (ii) Working substance is an ideal gas In real engine these condition cannot be fulfilled.

32.

No real engine can have an efficiency greater that of a carnot engine working between the same low temperatures. Why ?

Answer»

In carnot engine.

(i) There is absolutely no friction between the wall of cylinder and piston.

(ii) Working substance is an ideal gas In real engine these condition cannot be fulfilled.

Discussion

33.	If `HA+NaOH rarr NaA+H_(2)O" "DeltaH=-12kcal` and `HB+NaOH rarr NaB+H_(2)O" " DeltaH=-11kcal` then equimolar solution of which acid has higher `pH :`A. `HA`B. `HB`C. both have same `pH`D. information insufficient
Answer» Correct Answer - B HA will be stronger acid, so its solution will have lower pH.

Discussion

34.	The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the difference between CP and CV for 10 moles of an ideal gas.
Answer» C_P – C_V = nR = 10 × 4.184 J

34.

The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the difference between CP and CV for 10 moles of an ideal gas.

Answer»

C_P – C_V = nR

= 10 × 4.184 J

Discussion

35.	Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?
Answer» (–w) = p_ext (V₂–V₁) = 0 × (5 – 1) = 0 For isothermal expansion q = 0 By first law of thermodynamics q = ΔU + (–w) ⇒ 0 = ΔU + 0 so ΔU = 0

35.

Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?

Answer»

(–w) = p_ext (V₂–V₁) = 0 × (5 – 1) = 0

For isothermal expansion q = 0

By first law of thermodynamics

q = ΔU + (–w)

⇒ 0 = ΔU + 0 so ΔU = 0

Discussion

36.	Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:a) What is the final pressure of the gas in A and B? b) What is the change in internal energy of the gas? c) What is the change in the temperature of the gas? d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?
Answer» a) 0.5 atm b) Zero c) Zero d) No a) The volume available to the gas is doubled as soon as the stopcock between cylinders A and B is opened. Since volume is inversely proportional to pressure, the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the pressure in each cylinder will be 0.5 atm. b) The internal energy of the gas can change only when work is done by or on the gas. Since in this case no work is done by or on the gas, the internal energy of the gas will not change. c) Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all. d) The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in nonequilibrium states, they do not lie on the P-V-T surface of the system.

36.

Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:a) What is the final pressure of the gas in A and B? b) What is the change in internal energy of the gas? c) What is the change in the temperature of the gas? d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?

Answer»

a) 0.5 atm

b) Zero

c) Zero

d) No

a) The volume available to the gas is doubled as soon as the stopcock between cylinders A and B is opened. Since volume is inversely proportional to pressure, the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the pressure in each cylinder will be 0.5 atm.

b) The internal energy of the gas can change only when work is done by or on the gas. Since in this case no work is done by or on the gas, the internal energy of the gas will not change.

c) Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.

d) The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in nonequilibrium states, they do not lie on the P-V-T surface of the system.

Discussion

37.	When a system is taken from state `B` to state `A` along path `BDA` as shown in figure below, `60 J` of heat flows out of the system and `10J` of work is doen on path `ACB` , then the heat corresponding to the processes `AC` and `BC` is respectively. A. `q_(AC)=-20J` & `q_(BC)=-50J`B. `q_(AC)=-20 & q_(BC)=50J`C. `q_(AC)=20J&q_(DB)=50J`D. `q_(AC)=20J & q_(BC)=-50J`
Answer» Correct Answer - 4 ln ACB process, AC process is isochoric, so `W_(AC)=0`. So, `" "DeltaE_(BC)=q_(ac)+W_(ac)` `-30=q_(BC)+(20)` `q_(BC)=-50J` Now, `" "q_(AB)+q_(AC)+q_(CB)` `70=q_(AC)+50(` since for path `BDA,E_(A)-E_(B)+q+W=-60+10=-50` J and `W_(ACB)=-20J,` So, `q_(AB)=DeltaE_(AB)-W_(ACB)=50+20=70J)` `q_(AC)=20J`

Discussion

38.	The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound:(a) is always negative(b) is always positive(c) may be positive or negative(d) is never negative
Answer» (c) May be positive or negative.

Discussion

39.	Calculate the enthalpy change for the processCCl4(g) → C(g) + 4Cl(g)and calculate bond enthalpy of C-Cl in CCl4(g),ΔvapH° (CCl4) = 30.5 kJ mol-1ΔfH° (CCl4) = -135.5 kJ mol-1ΔaH°(c) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation. ΔaH°(Cl2) = 242 kJ mol-1
Answer» CCl₄(g) → C(g) + 4Cl(g) Bond enthalpy of C-Cl bond = 1/4 x Heat energy required to break four C-Cl bonds in CCl₄(g) = 1/4 x ΔH The ΔH for the reaction CCl₄(g) → C(g) + 4Cl(g) is the enthalpy of atomisation of CCl₄ Δ_rH° = Σ bond enthalpies reactants - Σ bond enthalpies products = 30.5 - (715 + 4 x 242) kJ = 30.5 - (715 + 968) kJ = (30.5 - 1683) kJ = -1652.5 kJ

39.

Calculate the enthalpy change for the processCCl4(g) → C(g) + 4Cl(g)and calculate bond enthalpy of C-Cl in CCl4(g),ΔvapH° (CCl4) = 30.5 kJ mol-1ΔfH° (CCl4) = -135.5 kJ mol-1ΔaH°(c) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation. ΔaH°(Cl2) = 242 kJ mol-1

Answer»

CCl₄(g) → C(g) + 4Cl(g)

Bond enthalpy of C-Cl bond

= 1/4 x Heat energy required to break four

C-Cl bonds in CCl₄(g)

= 1/4 x ΔH

The ΔH for the reaction CCl₄(g) → C(g) + 4Cl(g) is the enthalpy of atomisation of CCl₄

Δ_rH° = Σ bond enthalpies reactants - Σ bond enthalpies products

= 30.5 - (715 + 4 x 242) kJ

= 30.5 - (715 + 968) kJ

= (30.5 - 1683) kJ

= -1652.5 kJ

Discussion

40.	Enthalpy of sublimation of a substance is equal to(i) enthalpy of fusion + enthalpy of vapourisation(ii) enthalpy of fusion(iii) enthalpy of vapourisation(iv) twice the enthalpy of vapourisation
Answer» (i) enthalpy of fusion + enthalpy of vapourisation

Discussion

41.	Is it possible to increase the temperature of gas without giving it heat ?
Answer» Yes, it happens during an adiabatic process.

Discussion

42.	Calculate average molar that capacity at constant volume of gaseous mixture contained 2 mole of each of two ideal gases `A(C_(v,m)=(3)/(2)R)` and `B(C_(v,m)=(5)/(2)R) :`A. `R`B. `2R`C. `3R`D. `8R`
Answer» Correct Answer - B Average `C_(v,m)=(n_(1)C_(v,m_(1))+n_(2)C_(v,m_(2)))/(n_(1)+n_(2))=(2xx(3)/(2)R+2xx(5)/(2)R)/(2+2)=2R`

Discussion

43.	At `0^(@)C,DeltaH_(fus)^(@)=6kJ //mol`, change of entropy for freezing of one mole of ice will be `:`A. `oo`B. 0C. `21.98J//mol-K`D. `13.6J//mol-K`
Answer» Correct Answer - C `DeltaS=(DeltaH_(f)^(@))/(T_(f))=(6000)/(273)=21.97J//moll-K`

Discussion

44.	Given a. `NH_(3)(g)+3CI_(2)(g)rarr NCI_(3)(g)+3HCI(g),DeltaH_(1)` b. `N_(2)(g)+3H_(2)(g)rarr 2NH_(3)(g),DeltaH_(2)` c. `H_(2)(g)+CI_(2)(g)rarr 2HCI(g),DeltaH_(3)` Express the enthalpy of formation of `NCI_(3)(g)(Delta_(f)H^(Theta))` in terms of `DeltaH_(1), DeltaH_(2)`,and `DeltaH_(3)`.A. `DeltaH_(f)=-DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`B. `DeltaH_(f)=DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`C. `DeltaH_(f)=DeltaH_(1)-(DeltaH_(2))/(2)-(3)/(2)DeltaH_(3)`D. None
Answer» Correct Answer - 1 `(1)/(2)N_(2)(g)+(3)/(2)Cl_(2)(g)rarr NCl_(3)(g)` `DeltaH=-DeltaH_(1)+(DeltaH_(2))/(2)-(3)/(2)DeltaH_(2)`

Discussion

45.	In the isothermal reversible compression of `52.0m` mol of a perfect gas at `260 K`, the volume of the gas is reduced to one`-`third of its initial value. Calculate `w` of this provess `y`.
Answer» Correct Answer - 2 `x=-nRT ln(V_(f))/(V_(l)),V_(f)=1//3V` `nRT=(-52.0xx10^(-3)mol)xx(8.314JK^(-1)mol^(-1))xx(260K)=1.12bar(4)xx10^(2)J` `w=(1.12bar(4)xx10^(2)J)xxln.(1)/(3)=+123J`

Discussion

46.	It the following processes, identify the irreversible process `:`A. B. C. D.
Answer» Correct Answer - 4 Continuous graph can be plotted for reversible process not for irreversible process.

Discussion

47.	If 1 mole of an ideal gas expands isothermally at `37^(@)C` from 15 litres to 25 litres, the maximum work obtained is `:`A. `1316.8J`B. `6.43J`C. `8.57J`D. `2.92J`
Answer» Correct Answer - A `w_(max.iso)=-2.303nRTlog.(V_(2))/(V_(1))` `=-2.303xx1xx0.08211xx310xxlog.(25)/(15)` `=-58.61xx0.22=-12.87J` `( - ve` sign shows work of expansion i.e. work done by the system )

Discussion

48.	A liquid which is confined inside an adiabatic piston is suddently taken from state `-1` to state `-2` by a single stage irreversible process. If the piston comes to rest at point 2 as shown, then the enthalpy change for the process will be `:` A. `DeltaH=(2gammaP_(0)V_(0))/(gamma-1)`B. `DeltaH=(3gammaP_(0)V_(0))/(gamma-1)`C. `DeltaH=-P_(0)V_(0)`D. None of these
Answer» Correct Answer - 3 Since liquid is expanding against external pressure `P_(0)` hence work done `w=-P_(0)(4V_(0)-V_(0))=-3P_(0)V_(0)` `DeltaU=w=-3P_(0)V_(0)` `rArr" "DeltaH=DeltaU+P_(2)V_(2)-P_(1)V_(1)=-3P_(0)V_(0)+4P_(0)V_(0)-2P_(0)V_(0)`

Discussion

49.	In which of the following changes at constant pressure, is work done by system on surrounding ?A. `H_(2)O(g) rarr H_(2)O(l)`B. `H_(2)O(s) rarr H_(2)O(l)`C. `H_(2)O(s) rarr H_(2)O(g)`D. `H_(2)O(g) rarr H_(2)O(s)`
Answer» Correct Answer - 3 Work will be doen by system in case of expansion `i.e. H_(2)O(s) rarr H_(2)O(g)`

Discussion

50.	A sample of oxygen gas expands its volume from `3L` to `5L` against a constant pressure of `3` atm. If work done during expansion be used to heat `10 ` mole of water initially present at `290K` , its finally temperature will be `(` specific heat capacity of water `=4.18 J//k-g) :`A. `292.0K`B. `298.0K`C. `290.8K`D. `293.7K`
Answer» Correct Answer - C `P(V_(2)-V_(1))=mSDeltaT` `3(2)xx101.3=180xx4.18xx(T_(f)-290)` `T_(f)=290.8K`

Discussion

Explore topic-wise InterviewSolutions in .

Calculate the enthalpy change when `50 mL` of `0.01 M Ca(OH)_(2)` reacts with `25mL` of `0.01 M HCI`. Given that `DeltaH^(Theta)` neutralisaiton of strong acid and string base is `140 kcal mol^(-1)`A. `14 cal`B. `35 cal`C. `10 cal`D. `7.5 cal`

If an average person jogs, hse produces 14.5 × 103 cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 103 cal for evaparation) is(a) 0.25 kg(b) 2.25 kg(c) 0.05 kg(d) 0.20 kg

For the reaction at `25^(@)C, C_(2)O_(4)(l) rarr 2XO_(2)(g)` `Delta H =2.1 kcal` and `DeltaS=20 cal K^(-1)` . The reaction would be `:`A. spontaneousB. non`-` spontaneousC. at equilibriumD. unpredictable

A reaction has `DeltaH=-33KJ` and `DeltaS=-58(J)/(K)` . This reaction would be:A. spontaneous at all temperatureB. non`-` spontaneous at all temperaturesC. spontaneous above a certain temperature onlyD. spontaneous below a certain temperature only

During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is

The enthalpy of combustion of graphite is 393.4 kJ. Calculate(a) The amount of graphite needed to produce 196.7 KJ of heat.(b) The number of moles of CO2 formed when 196.7 KJ of heat is produced.(c) The volume of oxygen required at S.T.P. to from 24.0 g of graphite in this process.

What is the final temperature of `0.10` mole monoatomic ideal gas that performs `75 cal` of work adiabatically.if the initial temperature is `227 ^(@)C` ( use `R=2 cal //K-mol)`A. `250K`B. `300K`C. `350K`D. `750K`

At 298 K. Kp for the reaction N2O4 (g) ⇄ 2NO2 (g) is 0.98. Predict whether the reaction is spontaneous or not.

A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in Fig. 6.1. What will be the value of ΔH for the cycle as a whole?

18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol–1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?

Which quantity out of ΔrG and ΔrGΘ will be zero at equilibrium?

Identify the state functions and path functions out of the following :enthalpy, entropy, heat, temperature, work, free energy.

One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?

The molar enthalpy of vapourisation of acetone is less than that of water. Why?

Match the following :(i) Entropy of vapourisation(a) decreases(ii) K for spontaneous process(b) is always positive(iii) Crystalline solid state(c) lowest entropy(iv) ΔU in adiabatic expansion of ideal gas(d) ΔHvap/Tb

Consider the reaction at `300K` `C_(6)H_(6)(l)+(15)/(2)O_(2)(g)rarr 6CO_(2)(g)+3H_(2)O(l),DeltaH=-3271`kJ What is `DeltaU` for the combustion of `1.5` mole of benzene at `27^(@)C` ?A. `-3267.25kJ`B. `-4900.88kJ`C. `-4906.5kJ`D. `-3274.75kJ`

In an exothermic reaction, heat is evolved, and system loses heat to the surrounding. For such system(i) qp will be negative(ii) ΔrH will be negative(iii) qp will be positive(iv) ΔrH will be positive

A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig. 12.11. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R)AB : constant volumeBC : constant pressureCD : adiabaticDA : constant pressure

If a air is a cylinder is suddenly compressed by a piston. What happens to the pressure of air?

Air pressure in a car tyre increases during driving. Explain.

What is the ratio of slopes of P-V graphs of adiabatic and isothermal process?

Explain the Heat of formation acetylene is +54 kcal.

No real engine can have an efficiency greater that of a carnot engine working between the same low temperatures. Why ?

If `HA+NaOH rarr NaA+H_(2)O" "DeltaH=-12kcal` and `HB+NaOH rarr NaB+H_(2)O" " DeltaH=-11kcal` then equimolar solution of which acid has higher `pH :`A. `HA`B. `HB`C. both have same `pH`D. information insufficient

The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the difference between CP and CV for 10 moles of an ideal gas.

Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?

The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound:(a) is always negative(b) is always positive(c) may be positive or negative(d) is never negative

Calculate the enthalpy change for the processCCl4(g) → C(g) + 4Cl(g)and calculate bond enthalpy of C-Cl in CCl4(g),ΔvapH° (CCl4) = 30.5 kJ mol-1ΔfH° (CCl4) = -135.5 kJ mol-1ΔaH°(c) = 715.0 kJ mol-1,where ΔaH° is enthalpy of atomisation. ΔaH°(Cl2) = 242 kJ mol-1

Enthalpy of sublimation of a substance is equal to(i) enthalpy of fusion + enthalpy of vapourisation(ii) enthalpy of fusion(iii) enthalpy of vapourisation(iv) twice the enthalpy of vapourisation

Is it possible to increase the temperature of gas without giving it heat ?

Calculate average molar that capacity at constant volume of gaseous mixture contained 2 mole of each of two ideal gases `A(C_(v,m)=(3)/(2)R)` and `B(C_(v,m)=(5)/(2)R) :`A. `R`B. `2R`C. `3R`D. `8R`

At `0^(@)C,DeltaH_(fus)^(@)=6kJ //mol`, change of entropy for freezing of one mole of ice will be `:`A. `oo`B. 0C. `21.98J//mol-K`D. `13.6J//mol-K`

In the isothermal reversible compression of `52.0m` mol of a perfect gas at `260 K`, the volume of the gas is reduced to one`-`third of its initial value. Calculate `w` of this provess `y`.

It the following processes, identify the irreversible process `:`A. B. C. D.

If 1 mole of an ideal gas expands isothermally at `37^(@)C` from 15 litres to 25 litres, the maximum work obtained is `:`A. `1316.8J`B. `6.43J`C. `8.57J`D. `2.92J`

In which of the following changes at constant pressure, is work done by system on surrounding ?A. `H_(2)O(g) rarr H_(2)O(l)`B. `H_(2)O(s) rarr H_(2)O(l)`C. `H_(2)O(s) rarr H_(2)O(g)`D. `H_(2)O(g) rarr H_(2)O(s)`