Please Explain:
Asked by nitishkrnehu09
| 1st Jan, 2018,
03:50: AM
Expert Answer:
Correct answer is option (C)
As for reversible adiabatic process change in entropy is less than zero.
For irreversible adiabatic process the change in entropy is greater than zero.
So option (A) is incorrect as Q → R is reversible process.
option (d) is also incorrect as ΔSPQ
As it should be greater than zero.
In case of option (B) P → R is reversible process. Therefore ΔSPR is not greater than zero.
So it is also incorrect.
Answered by Varsha
| 1st Jan, 2018,
04:52: PM
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Asked by nitishkrnehu09 | 1st Jan, 2018, 03:50: AM
Expert Answer:
Correct answer is option (C)
As for reversible adiabatic process change in entropy is less than zero.
For irreversible adiabatic process the change in entropy is greater than zero.
So option (A) is incorrect as Q → R is reversible process.
option (d) is also incorrect as ΔSPQ
As it should be greater than zero.
In case of option (B) P → R is reversible process. Therefore ΔSPR is not greater than zero.
So it is also incorrect.
Answered by Varsha | 1st Jan, 2018, 04:52: PM
Related Videos
- write a relation between delta G and Q and define the meaning of each term and answer the following a) why a reaction proceeds forward when Q< K and no net reaction occurs when Q = K b) explain the effect of increase in pressure in terms of reaction quotient Q for the reaction CO ⇆ CH4(g) + H2O(g) (ii) for the reaction, N2(g) +3H2(g)⇆ 2NH(g) at 400K , Kp = 41, find the value Kp for the reaction 2N2(g) +6H2(g) ⇆ 4NH3(g)
- Give the relationship between Gibbs energy change and equilibrium constant?
- What is the relation between ?
- At what temperature the entropy of a perfectly crystalline substance is zero?
- For a reaction,
Determine the temperature above which reaction will be spontaneous.
- Calculate the entropy of vaporization of water if its enthalpy of vaporization is 186.5 kJmol-1?
- For a water gas reaction at 1000°C the standard Gibb’s energy change is -8.1 kJ mol-1. Calculate the value of equilibrium constant?
- Calculate the standard Gibbs energy change for the following reaction:
Given that for Cu2+(aq) and Zn2+(aq) as 65 kJ mol-1 and -147.2 kJ mol-1 respectively.
- For the reaction:
Calculate the temperature at which Gibbs energy change ΔG is equal to zero. Predict the nature of the reaction at this temperature and above it.
- Calculate the standard energy Gibbs energy change for the formation of propane, C3H8 (g) at 298 K. Given that For propane = -103.85 kJmol-1S° mC3H8 (g) =270.2 JK-1 mol-1, S° m H2 (g) = 130.68 Jk-1mol-1 and S° m C(graphite) = 5.74 JK-1 mol-1
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