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  • Calculate the entropy change(J/K) when  of an ideal gas expands isothermally and reversibly from an

Calculate the entropy change(J/K) when 2\:\:mol of an ideal gas expands isothermally and reversibly from an initial volume of 1\:dm^{3}\:\:to\:\:10\:dm^{3} at 298\:\:K.

(Response should be upto one decimal digit like 78.5)

Option: 1

122.5


Option: 2

149.6


Option: 3

38.2


Option: 4

183.6


Answers (1)

best_answer

As we have learnt,

Entropy And Change In Entropy -

ENTROPY

It is a thermodynamic state quantity which is used to measure disorder or randomness of the molecules in a system The disorder or randomness in a system is measured in terms of entropy (S). The absolute value of 'S' is not determined so mostly change In entropy \DeltaS is measured.

Randomness \propto Entropy

It is a state function that depends only on the initial and final state of the system that is, it is independent of the path used in going from the initial to the final state.

\Delta \mathrm{S}=\mathrm{S}_{\mathrm{Final}}-\mathrm{S}_{\mathrm{Initial}}

For a general chemical reaction at 298K and 1 atm:

\begin{array}{l}{\mathrm{m}_{1} \mathrm{P}+\mathrm{m}_{2} \mathrm{Q} \rightarrow \mathrm{n}_{1} \mathrm{R}+\mathrm{n}_{2} \mathrm{S}} \\ {\Delta \mathrm{S}^{\circ}=\left[\left(\mathrm{n}_{1} \mathrm{S}_{\mathrm{R}}^{\circ}+\mathrm{n}_{2} \mathrm{S}_{\mathrm{s}}^{\circ}\right)-\left(\mathrm{m}_{1} \mathrm{S}_{\mathrm{P}}^{\circ}+\mathrm{m}_{2} \mathrm{S}_{\mathrm{Q}}^{\circ}\right)\right]} \\ {\Delta \mathrm{S}^{\circ}=\sum \mathrm{S}_{\mathrm{P}}^{\circ}-\sum \mathrm{S}_{\mathrm{R}}^{\circ}}\end{array}

It is an extensive property and a state function that depends on state variables like T, P, V and n which govern the state of a system.

Entropy and Temperature

Entropy increases with the increase of temperature as it is associated with the molecular motion which Increases with the increase of temperature due to an Increase the average kinetic energy of the molecules. The entropy of a perfectly ordered Crystalline substance is taken as zero at Zero Kelvin (0K) . It is third law of Thermodynamics. However in case of N2O, NO, LO, Solid CI2 etc. the value of entropy is not found to be zero at O Kelvin also.

We have given,

Initial volume, V1=1dm^{3}

Final volume, V2=10dm^{3}

Temperature, 298\:k

Number of moles, n = 2

Change in entropy in the reversible isothermal process is given as,  \Delta S=nR\:In\:\frac{V_{2}}{V_{1}}

\Delta S=2\times8.314\:In\:\frac{10}{1}

\Delta S=38.2J/K

Posted by

shivangi.bhatnagar

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