Question 1

Which of the following processes exhibits the decrease in internal energy of a system?

Accepted Answer

According to the first law of thermodynamics, U = q + w. For an isothermal process, the temperature remains constant ( T = 0). Since the internal energy of an ideal gas depends only on temperature, U = 0 for both isothermal expansion and compression. For an adiabatic process, there is no heat exchange with the surroundings, so q = 0. The first law becomes U = w. During adiabatic expansion, work is done by the gas on the surroundings, meaning w During adiabatic compression, work is done on the ga

Question 2

Calculate the change in internal energy if an ideal gas expands from 0.5\ dm ^3 to 3.5\ dm ^3 at constant pressure 2 bar by absorbing 1800 J heat energy.

Accepted Answer

Given V_1 = 0.5\ dm ^3, V_2 = 3.5\ dm ^3, P = 2\ bar , and heat absorbed q = 1800\ J . Work done during expansion at constant pressure is given by W = -P(V_2 - V_1). W = -2 (3.5 - 0.5) = -6\ bar dm ^3 Since 1\ bar dm ^3 = 100\ J , we have W = -600\ J . According to the first law of thermodynamics, U = q + W. Substituting the values, U = 1800 - 600 = 1200\ J . Answer: 1200 J.

Question 3

The enthalpy of combustion of C, H _2, and CH _4 are -390, -285 and -890 kJ/mol. Find the enthalpy of formation of methane.

Accepted Answer

The required reaction for the formation of methane is: C(s) + 2 H _2 (g) CH _4 (g) The enthalpy of formation can be calculated using the enthalpies of combustion of the reactants and products: H_f( CH _4) = H_c( C ) + 2 H_c( H _2) - H_c( CH _4) Substituting the given values: H_f( CH _4) = -390 + 2(-285) - (-890) H_f( CH _4) = -390 - 570 + 890 H_f( CH _4) = -960 + 890 = -70 kJ/mol Answer: -70 kJ

Question 4

Identify the law for the statement "Overall the enthalpy change for a reaction is equal to sum of enthalpy changes of individual steps in the reaction"

Accepted Answer

Hess's law of constant heat summation states that the total enthalpy change for a chemical reaction is independent of the path by which the reaction occurs. If a reaction takes place in several steps, the overall enthalpy change is equal to the sum of the enthalpy changes of the individual steps. This is because enthalpy is a state function, meaning it depends only on the initial and final states of the system, not on the path taken. Answer: Hess's law

Question 5

If 2 mole of an ideal gas expand isothermally and reversibly at 27^ from 1 dm ^3 to 1\ m ^3 calculate work done? [R = 8.314\ J K ^ -1 mol ^ -1 ]

Accepted Answer

Given: n = 2 mol T = 27^ C = 300 K V_1 = 1 dm ^3 V_2 = 1 m ^3 = 1000 dm ^3 R = 8.314 J K ^ -1 mol ^ -1 The work done in a reversible isothermal expansion is given by: W = -2.303 nRT _ 10 ( V_2 V_1 ) Substituting the given values: W = -2.303 2 8.314 300 _ 10 ( 1000 1 ) W = -2.303 600 8.314 3 W = -34464.8 J W = -34.46 kJ Answer: -34.46 kJ

Question 6

Which from following is NOT a state function ?

Accepted Answer

State functions are properties whose values depend only on the state of the system and are independent of the path followed to reach that state. Pressure, volume, and temperature are state functions because they depend only on the initial and final states of the system. Work and heat are path functions because their values depend on the specific path taken to go from the initial state to the final state. Therefore, work is not a state function. Answer: Work

Question 7

Calculate the work done for the following reaction at 27\,^ C C _2 H _ 4(g) + H _ 2(g) C _2 H _ 6(g) (R = 8.314\, JK ^ -1 mol ^ -1 )

Accepted Answer

The given reaction is C _2 H _ 4(g) + H _ 2(g) C _2 H _ 6(g) Change in number of moles of gaseous species, n_g = n_p - n_r n_g = 1 - (1 + 1) = -1 Work done, W = - n_g RT W = -(-1) 8.314 (27 + 273) W = 1 8.314 300 W = 2494.2 J Answer: 2494.2 J

Question 8

Calculate standard Gibbs energy for any gaseous reaction if values of K_p and R respectively are 3.5 10^ 17 ; 8.314\, JK ^ -1 mol ^ -1

Accepted Answer

The standard Gibbs free energy change is given by the relation: G^ = -RT K_p = -2.303 RT K_p Assuming standard temperature T = 298\, K : G^ = -2.303 8.314 298 (3.5 10^ 17 ) G^ = -5705.8 ( 3.5 + 17) G^ = -5705.8 (0.544 + 17) G^ = -5705.8 17.544 G^ = -100102\, J mol ^ -1 -100.1\, kJ mol ^ -1 Answer: -100.1\, kJ mol ^ -1

Question 9

An ideal gas expands from initial volume 5\ dm ^3 to 15\ dm ^3 against a constant external pressure of 2 atm. What is the work done by the gas ?

Accepted Answer

The work done by a gas during expansion against a constant external pressure is given by the formula: W = -P_ ext V Given: P_ ext = 2\ atm V_i = 5\ dm ^3 = 5\ L V_f = 15\ dm ^3 = 15\ L Change in volume: V = V_f - V_i = 15 - 5 = 10\ L Substituting the values into the formula: W = -2\ atm 10\ L = -20\ L \ atm To convert the work done into Joules, use the conversion factor 1\ L \ atm = 101.32\ J : W = -20 101.32\ J = -2026.4\ J Answer: -2026.4 J

Question 10

A gas absorbs 120 J of heat and expands by 300 10^ -6 \ m ^3 against a constant external pressure of 2 10^5\ Nm ^ -2 . What will be the change in internal energy of the system ?

Accepted Answer

Heat absorbed by the gas, q = +120\ J Work done by the gas during expansion against constant external pressure is given by W = -P_ ext V Substituting the given values: W = -(2 10^5\ Nm ^ -2 ) (300 10^ -6 \ m ^3) W = -60\ J According to the first law of thermodynamics: U = q + W U = 120\ J - 60\ J = 60\ J

Question 11

Identify from following, the relation between standard free energy change in a reaction and equilibrium constant K_c.

Accepted Answer

The relation between the free energy change and the reaction quotient is given by G = G^ + RT Q. At equilibrium, the free energy change G = 0 and the reaction quotient Q = K_c. Substituting these values into the equation, we get 0 = G^ + RT K_c. Rearranging the terms yields - G^ = RT K_c. Answer: - G^ = RT K_c

Question 12

Calculate the standard enthalpy change for the following reaction, 2 C _2 H _6 (g) + 7 O _2 (g) 4 CO _2 (g) + 6 H _2 O (l) Given, _f H^ ( C _2 H _6) = -85\ kJ mol ^ -1 _f H^ ( CO _2) = -390\ kJ mol ^

Accepted Answer

The standard enthalpy of reaction is calculated using the formula: _r H^ = _f H^ ( products ) - _f H^ ( reactants ) For the given reaction: _r H^ = [4 _f H^ ( CO _2) + 6 _f H^ ( H _2 O )] - [2 _f H^ ( C _2 H _6) + 7 _f H^ ( O _2)] Substituting the given values: _r H^ = [4(-390) + 6(-285)] - [2(-85) + 7(0)] _r H^ = [-1560 - 1710] - [-170] _r H^ = -3270 + 170 = -3100\ kJ Answer: -3100\ kJ

Question 13

Which of the following is the correct expression of the first law of thermodynamics for adiabatic processes ?

Accepted Answer

According to the first law of thermodynamics, the change in internal energy is given by U = Q + W. For an adiabatic process, there is no heat exchange between the system and the surroundings, which means Q = 0. Substituting Q = 0 into the first law equation, we get U = W. Multiplying both sides by -1, this can be written as - U = -W. Answer: - U = -W

Question 14

1 mole of an ideal gas is compressed isothermally and reversibly from initial pressure x kPa to final pressure 2x kPa at 300 K. Find the work done (R = 8.314\ J K ^ -1 mol ^ -1 )

Accepted Answer

For an isothermal reversible process, the work done is given by: W = nRT ( P_2 P_1 ) Substituting the given values: n = 1 mol T = 300 K P_1 = x kPa P_2 = 2x kPa R = 8.314 J K^ -1 mol^ -1 W = 1 8.314 300 ( 2x x ) W = 2494.2 (2) W = 2494.2 0.693 W 1729 J Answer: 1729 J

Question 15

Which from following is correct expression of first law of thermodynamics for isothermal process?

Accepted Answer

According to the first law of thermodynamics, the change in internal energy is given by U = Q + W. For an isothermal process, the temperature of the system remains constant, so T = 0. Since the internal energy of an ideal gas depends only on temperature, the change in internal energy is zero, U = 0. Substituting U = 0 into the first law equation gives: 0 = Q + W W = -Q Answer: W = - Q

Question 16

For a certain reaction, H ^ = 40\ kJ and S ^ = 80\ JK ^ -1 . Find the temperature so that G ^ = 0

Accepted Answer

Using the Gibbs free energy equation: G^ = H^ - T S^ For G^ = 0, we get: T = H^ S^ Given H^ = 40\ kJ = 40000\ J and S^ = 80\ J K ^ -1 . Substituting the values: T = 40000 80 = 500\ K Answer: 500\ K

Question 17

Calculate H for the following reaction at 300\ K 2 C _ (s) + 3 H _ 2(g) C _2 H _ 6(g) if U for the reaction is -80\ kJ ( R = 8.314\ JK ^ -1 mol ^ -1 )

Accepted Answer

The given reaction is 2 C _ (s) + 3 H _ 2(g) C _2 H _ 6(g) Change in number of gaseous moles, n_g = n_ p(g) - n_ r(g) = 1 - 3 = -2 Given U = -80\ kJ T = 300\ K R = 8.314\ J K ^ -1 mol ^ -1 = 8.314 10^ -3 \ kJ K ^ -1 mol ^ -1 Using the relation H = U + n_g RT H = -80 + (-2) 8.314 10^ -3 300 H = -80 - 4.9884 H = -84.9884\ kJ -85.00\ kJ Answer: -85.00\ kJ

Question 18

Which of the following processes is an endothermic ?

Accepted Answer

Mixing of aqueous NaOH and HCl is a neutralization reaction, which is highly exothermic. Formation of ammonia by the Haber process is an exothermic reaction as heat is released. Melting of ice requires the absorption of heat from the surroundings to overcome the intermolecular forces of attraction between water molecules. Thus, it is an endothermic process. Freezing of liquid water releases heat to the surroundings, making it an exothermic process. Answer: Ice melts at room temperature.

Question 19

Calculate the standard enthalpy of combustion of carbon monoxide if _f H^ ( CO ) = -110 kJ mol ^ -1 _f H^ ( CO _2) = -393 kJ mol ^ -1

Accepted Answer

The combustion reaction of carbon monoxide is given by: CO(g) + 1 2 O _2 (g) CO _2 (g) The standard enthalpy of combustion is calculated using the standard enthalpies of formation: _c H^ ( CO ) = _f H^ ( products ) - _f H^ ( reactants ) _c H^ ( CO ) = _f H^ ( CO _2) - [ _f H^ ( CO ) + 1 2 _f H^ ( O _2) ] Since the standard enthalpy of formation of O _2 in its standard state is zero: _c H^ ( CO ) = -393 - (-110 + 0) _c H^ ( CO ) = -393 + 110 = -283 kJ mol ^ -1 Answer: -283 kJ mol ^ -1

Question 20

Calculate the quantity of heat released from system when 2 moles of and ideal gas compressed isothermally from volume 25 dm ^3 to 10 dm ^3 at constant external pressure 4 bar.

Accepted Answer

For an isothermal process of an ideal gas, the change in internal energy U = 0. From the first law of thermodynamics: U = q + w q = -w The work done during irreversible compression at a constant external pressure is given by: w = -P_ ext V = -P_ ext (V_2 - V_1) Given: P_ ext = 4 bar V_1 = 25 dm ^3 V_2 = 10 dm ^3 Substituting the values: w = -4 (10 - 25) = 60 bar dm ^3 Since 1 bar dm ^3 = 100 J : w = 60 100 J = 6000 J = 6.0 kJ Therefore, q = -6.0 kJ . The quantity of heat released from the system

Thermodynamics (C) — Previous Year Questions & Solutions