AIIMS Chemistry Thermodynamics Class 11 Questions

C + O₂(g) ⟶ CO₂ ...... (i); ΔH = -393 kJ mol⁻¹ H₂ + 1/2 Ω₂ ⟶ H₂O ...... (ii); ΔH = -287.3 kJ mol⁻¹ 2CO₂ + 3H₂O ⟶ C₂H₅OH + 3O₂ ...... (iii); ΔH = 1366.8 kJ mol⁻¹ Find the standard enthalpy of formation of C₂H₅OH(l)

In isolated system, find the condition for spontaneous reaction:

The entropy for a system can be calculated as: ΔS = nCᵥ ln(T₂/T₁) + nR ln(V₂/V₁)……(I)

In adiabatic expansion, work is done at the cost of internal energy. Hence, the temperature of system is decrease.

The state function is dependent only on the initial and final state of the system. Intensive properties are those which are independent of the amount of the substance. For example temperature, melting point, boiling point etc. The internal energy (U) is a state function.

According to the 1ˢᵗ law of thermodynamics ΔU = q + w If, w = 0 ΔU = q q depends on the reaction path travelled and thus not a state function.

The equation for standard Gibbs free energy given below. ΔG = ΔG° + RTlnQ…(1) If ΔG = 0 and Q = K then the above equation at equilibrium becomes, ΔG° = −RTlnK …(2) Substituting equation (2) in equation (1), the new relation forms as, ΔG = −RTlnK + RTlnK ΔG = −RTln(K/Q)

The change in the enthalpy is,

The Boyle’s law states that pressure is inversely proportional to volume as P ∝ 1/V and temperature is constant. The graph for the same is given below. For adiabatic process, the equation is given as PVᵞ = constant So, the assertion is true but the reason is false.

At 25°C 1 mole of butane is heated then CO₂ and H₂O liquid is formed work done is :

From the Gibbs free energy, ΔG° = ΔH° - TΔS° For spontaneous reaction, ΔG° = 0 Now, ΔH° - TΔS° < 0 T = ΔH°/ΔS° Substitute the given value in the above expression. T > 179.1×10³/160.2 > 1118 K

The equation of Gibb's free energy is, ΔG = ΔG° + RT log Kc. At equilibrium, ΔG = 0. So, ΔG° = -RT log Kc.

At critical state, the compressibility factor (Z) for a real gas is equal to

One monatomic gas is expanded adiabatically from 2 L to 10 L at 1 atm external pressure, find ΔU (in atm L)?

Which of the following are extensive properties?

The volume (x) for 3 mole of gas can be calculated by using the expression shown below:

The expression for pressure ratio is, \( \frac{P_1}{P_2} = \frac{T_1}{T_2} \) Substitute the values in above formula. \( \frac{10 \text{ atm}}{15 \text{ atm}} = \frac{300 \text{ K}}{T_2} \) \( T_2 = 450 \text{ K} \)

The factor of ΔG values is important in metallurgy. The ΔG values for the following reactions at 800°C are given as S₂(g) + 2O₂(g) → 2SO₂(g); ΔG = -544kJ 2Zn(s) + S₂(g) → 2ZnS(s); ΔG = -293kJ 2Zn(s) + O₂(g) → 2ZnO(s); ΔG = -480kJ The ΔG for the reaction, 2ZnS(g) + 3O₂(g) → 2ZnO(g) + 2SO₂(g) will be

The temperature in process AB is constant so, volume increases and in the process BC entropy is same with decreased T which increases the volume.

ΔG versus T plot in the Ellingham’s diagram slopes downward for the reaction

The total entropy (universe) is equal to the sum of entropy of system and entropy of surroundings. ΔS_universe = ΔS_system + ΔS_surroundings ……(I) ΔS_surroundings = ΔH_surroundings/T = -ΔS_system/T Therefore, the equation (I) will become as: ΔS_universe = ΔS_system + (-ΔH_surroundings/T) The above equation shows that for a spontaneous system, the entropy of universe is continuous increasing.

C₃H₆ + H₂ → C₃H₈ ΔH₁ = -224 C₃H₈ + 5O₂ → 3CO₂ + 4H₂O ΔH₂ = -2027 H₂ + ½ O₂ → H₂O ΔH₃ = -282 Calculate the combustion of propene

Calculate work done

The expression for entropy change is given as, ΔS = nCₚ ln T₂/T₁ ΔS = 7/2 × R × ln 800/400 ΔS = 40 J/K

By first law of thermodynamics in case free expansion the change in internal energy is equal to zero. The work done in case of free expansion is zero. Thus, both assertion and reason is correct but they are not related to each other.

The properties which depend on the quantity of any species are known as extensive property. Therefore, volume and energy are extensive properties.

Assertion: Addition of Q and w give ΔU Reason: Addition of two path function can not give state function.

When ideal gas expands from initial state to final state, the work done will be high in which number of steps are higher. The assertion and reason are not related.

Assertion: Second ionization enthalpy will be higher than the first ionization enthalpy.

In case of isothermal process involving an ideal gas Gibbs and Helmbaltz free energy charges are equal. Thus assertion correct and its reason is also correct as isothermal process for an ideal gas both ΔE and ΔH are zero.

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choice. Assertion (A) Black body is an ideal body that emits and absorb radiations of all frequencies. Reason (R) The frequency of radiations emitted by a body goes from lower frequency to higher frequency with an increase in temperature.

Assertion: A reaction which is spontaneous and accompanied by decrease of randomness must be exothermic. Reason: All exothermic reactions are accompanied by decrease of randomness.

The factor of ΔG values is important in metallurgy. The ΔG values for the following reactions at 800°C are given as S₂(g) + 2O₂(g) → 2SO₂(g); ΔG = −544 kJ 2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g) will be

ΔG° versus T plot the Ellingham’s diagram slopes downward for the reaction

Which thermodynamic parameter is not a state function?

Assertion: For an isolated system, q is zero. Reason: In an isolated system, change in U and V is zero.

Assertion: Entropy of system increases for a spontaneous reaction. Reason: Enthalpy of reaction always decreases for spontaneous reaction.

Assertion: A process is called adiabatic if the system does not exchange heat with the surroundings. Reason: It does not involve increase or decrease in temperature of the system.

At equilibrium which is correct?

For adiabatic process, which is correct?

Which of the following is not a thermodynamic function?

Which of the following is intensive property?

Assertion : Entropy is always constant for a closed system. Reason : Closed system is always reversible.

Match List I with List II and select the correct answer using the codes given below the lists:

The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion (α) of a gas i.e., α = (1/V)(∂V/∂T)P. For an ideal gas α is equal to

In an isobaric process, when temperature changes from T₁ to T₂, ΔS is equal to

In P versus V graph, the horizontal line is found in which ______ exists.

Calculate change in internal energy if ΔH = –92.2 kJ, P = 40 atm and ΔH/ΔV = –1 L.

ΔHₓₓₓₓ of a substance is 'x' and ΔHₓₓₓₓ is 'y', then ΔHₓₓₓₓₓₓ will be

ΔSₛᵤᵣ for an exothermic reaction is