Home
Class 12
CHEMISTRY
Melting point of a solid is x K and its ...

Melting point of a solid is x K and its latent heat of fusion is 600 cal `mol^(-1)`. The entropy change for fusion of 1 mol solid is 2 cal `mol^(-1)K^(-1)`. The value of x will be:

Promotional Banner

Similar Questions

Explore conceptually related problems

Melting point of a solid is xK and its latent heat of fusion is 600 cal mol^(-1) . The entropy changes for fusion of 1 mol solid is 2 cal mol^(-1) K^(-1) . The value of x will be

The melting point of a solid is 300K and its latent heat of fusion is 600 cal mol^(-1) . The entropy change for the fusion of 1 mole of the solid ( in cal K^(-1)) at the same tempertre would be :

The melting point of a solid is 300K and its latent heat of fusion is 600 cal mol^(-1) . The entropy change for the fusion of 1 mole of the solid ( in cal K^(-1)) at the same tempertre would be :

The melting point of a solid is 300K and its latent heat of fusion is 600 cal mol^(-1) . The entropy change for the fusion of 1 mole of the soli (in cal K^(-1) ) at the same temperature would be:

The thermodynamic property that measures the extent of molecular disorder is called entropy. Entropy change of phase transformation can be calculated using Trouton's formula (DeltaS = DeltaH//T) . In the reversible adiabatic process, however, DeltaS will be zero. the rise in temperature in isobaric and isochoric conditions is found to increase the randomness or entropy of the system. DeltaS = 2.303 C log (T_(1)//T_(2)) The melting point of a solid is 200K and its latent heat of fusion is 400cal mol^(-1) . The entropy changes for the fusion of 1 mole of the solid (in cal K^(-1)) at the same temperature would be

The thermodynamic property that measures the extent of molecular disorder is called entropy. Entropy change of phase transformation can be calculated using Trouton's formula (DeltaS = DeltaH//T) . In the reversible adiabatic process, however, DeltaS will be zero. the rise in temperature in isobaric and isochoric conditions is found to increase the randomness or entropy of the system. DeltaS = 2.303 C log (T_(1)//T_(2)) The melting point of a solid is 200K and its latent heat of fusion is 400cal mol^(-1) . The entropy changes for the fusion of 1 mole of the solid (in cal K^(-1)) at the same temperature would be

Latent heat of fusion of ice is 6kJ mol^(-1) . Calculate the entropy change in the fusion of ice.

Latent heat of fusion of ice is 6 kJ mol^(-1) . Calculate the entropy change in the fusion of ice.