In which of the following processes entropy increases?
A. A liquid evaporates to vapour.
B. Temperature of a crystalline solid lowered from 130 K to 0 K.
C.
D.
Choose the correct answer from the options given below: [2024]
A and C
A, B and D
A, C and D
C and D
(3)
A. Molecules in gaseous phase have greater freedom of movement, i.e.,have greater randomness than those in liquid phase, hence entropy increases.
B. On decreasing temperature of a solid, entropy decreases.
C. As number of gaseous products increases, randomness increases.
D. The products have larger number of gaseous particles, hence entropy increases.
For irreversible expansion of an ideal gas under isothermal condition, the correct option is [2021]
(4)
For reversible and irreversible expansion for an ideal gas under isothermal condition, , but i.e., is not zero for irreversible process.
For the reaction, , the correct option is [2020]
and
and
and
and
(4)
In the reaction, , the randomness decreases as 2 moles of are converted to 1 mole of ; thus, .
And this is an exothermic reaction, thus, .
In which case change in entropy is negative? [2019]
Evaporation of water
Expansion of a gas at constant temperature
Sublimation of solid to gas
(1)
If then
For a given reaction, and . The reaction is spontaneous at (Assume that and do not vary with temperature.) [2017]
all temperatures
(1)
For a spontaneous reaction, i.e.,
;
For a sample of perfect gas when its pressure is changed isothermally from to , the entropy change is given by [2016]
(2)
For an ideal gas undergoing reversible expansion, when temperature changes from to and pressure changes from to ,
For an isothermal process, so,
Consider the following liquid-vapour equilibrium.
Which of the following relations is correct? [2016]
(2)
This is Clausius--Clapeyron equation.
Which of the following statements is correct for the spontaneous adsorption of a gas? [2014]
is negative and, therefore should be highly positive.
is negative and therefore, should be highly negative.
is positive and therefore, should be negative.
is positive and therefore, should also be highly positive.
(2)
Using Gibbs'-Helmholtz equation,
During adsorption of a gas, entropy decreases i.e.,
For spontaneous adsorption, should be negative, which is possible when is highly negative.
For the reaction,
, cal at 300 K
Hence, is [2014]
2.7 kcal
- 2.7 kcal
9.3 kcal
- 9.3 kcal
(2)
Given, kcal, ,
kcal, T = 300 K
kcal
Again,
Given, kcal
On putting the values of and in the equation, we get
kcal