Explore topic-wise InterviewSolutions in .

No of moles, n = 2

temperature, T = 298K

let initial volume = V_i

∴ final volume, V_f = 3V_i

The work done for isothermal process is given by,

W = 2.303 nRT log₁₀ \((\frac{v_2}{v_1})\) 

where V₂ is the final volume and V₁ is the initial volume.

Here \(\frac{v_2}{v_1}\) = \(\frac{v_f}{v_i}\) = 3

∴ W = 2.303 × 2 × 8.314 × 298 × \({log^3_{10}}\)

W = 5.44kJ

Since, for isothermal process, Heat supplied is equal to work done by the gas, the total heat supplied is 5.44 kJ.

(i) Sudden bursting of the tube of bicycle tyre. 

(ii) Propagation of sound waves in air and other gases.

A gas can be liquid by pressure alone, only when its temperature is below its critical temperature.

No, a gas can be liquefied by pressure alone, only when temperature of gas is below its critical temperature.

Acc. to this, when the thermodynamic system A and B are separately in thermal equilibrium with a third thermodynamic system C, then the system A and B are in thermal equilibrium with each other also. 

For NO(g); Δ_rH^θ = +ve : unstable in nature

For NO₂(g); Δ_rH^θ = -ve : stable in nature.

ΔG° = -2.303 RT log Kc

= -2.303 x (8.314 mol^-1 K^-1) x 300 K log 10

= -2.303 x 8.314 x 300 x 1

= -5527 J mol^-1

= -5.527 kJ mol^-1

(a) ΔG° = -2.303 RT logK If K > 1, ΔG° will be less than zero because products formed are more than that of reactants, i.e., process is spontaneous in forward direction. 

(b) It is because heat energy is required to overcome activation energy. 

(c) ΔG = ΔH - TΔS, At low temperatures ΔH > TΔS whereas at high temperature TΔS >ΔH 

∴ ΔG decreases, i.e., becomes negative.

(i) CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(l);

Δ_rH° = -726 kJ mol^-1

(ii) C(g) + O₂(g) → CO₂(g);

Δ_cH° = -393 kJ mol^-1

(iii) H₂(g) + 1/2O₂(g) → H₂O(l);

Δ_fH° = -286 kJ mol^-1

We aim at

C + 2H₂ + 1/2 O₂ → CH₃OH

In order to get this thermochemical equation, multiply Eq. (ii) by 1 and Eq. (iii) by 2 and subtract Eq. (i) from their sum.

C + 2H₂ + 1/2O₂ → CH₃OH

ΔH = (-393) + 2(-286) - (-726)

= -393 - 572 + 726

= -965 + 726 = -239 kJ mol^-1

Thus, the heat of formation of CH₃OH is

ΔH_f = -239 kJ mol^-1

Δ_fH° NH₃(g) = {-(92.4)}/{2} = -46.2 kJ mol^-1

(i) When temperature is raised, disorder in molecules increases and therefore entropy increases.

(ii) Reactant is a solid and hence has low entropy. Among the products there are two gases and one solid so products represent a condition of higher entropy. 

(iii) Here 2 moles of H atoms have higher entropy than one mole of hydrogen molecule.

Process which takes place itself, without any external aid under the given condition is called spontaneous process. 

Example: Flow of heat from higher to lower temperature.

(i) ΔG = -ve, ΔS = +ve 

(ii) ΔG = -ve, ΔS = -ve (if ΔH > TΔS) 

(iii) ΔG = ΔS = -ve

(a)(i) Irreversible, 

(ii) Irreversible, 

(iii) Irreversible, 

(iv) Reversible 

(b) It is because no work is done, 

i.e., w = 0 w = -pext × ΔV = 0 × ΔV = 0 

In ΔU = q + w 

q = 0 because gas chamber is insulated ΔU = 0 + 0 = 0

Process which does not takes place itself or on its own but with the help of external aid under the given condition is called non-spontaneous process.

Example: Flow of heat energy from lower to higher temperatures.

(i) ΔrH° = -57.1 kJmol^-1 

(ii) H₂(g) + I₂(g) → 2HI(g) here ΔH = ΔU