Class 11th

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We know that the amount of work done =-p_ext? V

On substituting the values in the formula, we get,

-2bar* (50-10)L=-80Lbar

According to the described problem,1 LBar = 100J

Therefore, -80 L bar= (-80*100)= -8000J

= -8kJ, which is the amount of work done

The significance of the negative sign states that the work is done on the surroundings of the system. In the case of reversible expansion, the work done will be more.

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We can conclude from the figure that this change is a reversible change.

Now,

W= -2.303nRT log $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

But,  p₁V₁ = p₂V_{2 = $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$} = $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$ = $\frac{2}{1}$ =2

 W= -2.303nRT log $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$

 = -2.303 *8.314*1*298*log2

= -2.303 *8.314*298*0.3010J

= -1717.46J

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No, for the state of spontaneity, the enthalpy change is not the only criteria. Entropy also needs to be taken into account here.

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Throwing a stone from ground to roof

b) the reaction involved is a process where the energy decreases after the reaction. It can be represented as:In process b), potential energy/enthalpy change is a contributing factor to the spontaneity.

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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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