Chemistry NCERT Exemplar Solutions Class 11th Chapter Six

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When a process can be reversed by bringing an extremely small change in it, we call it a reversible process. The pressure-volume graph can be used to calculate the work done. The pressure is not constant, and changes in infinitesimal amounts as compression happens from initial volume V_i to the final volume V_f. The below graph depicts the work done with the shaded area.

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Assumption: Cylinder is filled with one mole gas, and the piston is frictionless. Let the pressure of gas inside be p and the volume of gas be V_ {I}.

Piston is moved towards the inside to make the external pressure (P_ {ext}) equal to p. Now, let us assume that this change takes place in a single step, hence, V is the final volume. The work done by the piston is depicted in the graph shown below by shading the area.

P_extΔV= AV₁   (V₁-V₂)

Chemistry NCERT Exemplar Solutions Class 11th Chapter six 
Standard molar enthalpy of formation is the enthalpy change for the formation of one mole of a compound from its most stable states or reference states. As per the given information in the question, the standard enthalpy for the given equation is – 572 kJ mol^–1

Now the enthalpy of formation for H₂O will be half the enthalpy of the value in the given equation. So now we can calculate that

? _fH? = $\frac{-572\mathrm{}\mathrm{K}\mathrm{J}}{2}$ = -286KJ/mol

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As per the information provided in the question, for one mole of CCl₄ (154 g), the heat of vaporisation required is 30.5 kJ/mol .

Hence for the vaporisation of 284 g of CCl₄, we require:

$\frac{284}{154}*30.5$  = 56.2 kJ

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Δ_rH? =ΣB.E _(reactant)-ΣB.E _(product)

 =B.E._H2 + BE_BR2 -2 *B.E>_HBr

= 435+192-2 (*368)

= 109KJmol^-1

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Molar enthalpy change for graphite (ΔH)

= enthalpy change for 1 g x molar mass of C = -20.7*12 = -2.48 x 102 kJ mol-1

Since the sign of ΔH = -ve, it is an exothermic reaction.

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Given that, Cv = heat capacity at constant volume,

Cp = heat capacity at constant pressure

 Difference between Cp and Cv is equal to gas constant (R).

.'. Cp – Cv = nR                                (where, n = no. of moles)

= 10 x 8.314 = 83.14J

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For water, molar heat capacity = 18 x Specific heat or Cp = 18 x c

But, specific heat,

C = 4.18 J g^-1 K^-1 Heat capacity,

Cp = 18 x 4.18 JK^-1  = 75.24 JK^-1

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During free expansion, external pressure is zero, so Work done, w = -pextΔV

= -0(5 – 1) = 0

Since the gas is expanding isothermally, therefore, q = 0

ΔU = q + w =0+0=0