Thermodynamics

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This is a Multiple Choice Questions as classified in NCERT Exemplar

Thermodynamics is not concerned about how and at what rate these energy transformations are carried out but is based on initial and final states of a system undergoing the change. Laws of thermodynamics apply only when a system is in

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This is a Short Answer Type Questions as classified in NCERT Exemplar

We know that the amount of work done =-pext? 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 amo

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This is a Short Answer Type Questions as classified in NCERT Exemplar

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

Now,

W= -2.303nRT log V 2 V 1

But,  p1V1 = p2V2 = V 2 V 1 = P 1 P 2 = 2 1 =2

 W= -2.303nRT log P 1 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 spont

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This is a Short Answer Type Questions as classified in NCERT Exemplar

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 infinitesim

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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

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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 eq

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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:

284 154 * 30.5  = 56.2 kJ

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