Answer
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Hint: The change in enthalpy in a system can be given as dH = dU + dW. Where dH is the change in enthalpy, dU is the change in internal energy of the system and dW is the change is work.
Complete step by step solution:
We know that the state function G is called Gibbs free energy of the system. The value of dG decided whether the reaction will be spontaneous or nonspontaneous.
- Here, we are given the formula of Gibbs free energy G which can be written as
G = H – TS ….(1)
Here, H is the enthalpy, T is the absolute temperature of the system and S is the entropy.
- So, now we need to find which quantities will be constant in the spontaneous process involving only PV work.
We can write the equation (1) in its differential form as
dG = dH – TdS
Now, dH is the enthalpy change and it is related with internal energy (U) and work (W) as
dH = dU + dW ….(2)
Now, we are given that work is of PV type only. So, we can write that dW = PdV. So, we can write the equation (2) as
dH = dU + pdV …(3)
Now, put equation (3) in equation (1), so we will get
dG = dU + PdV – TdS
Now, for a spontaneous process, dG < 0 and TdS > dU+PdV and VdP=SdT=0 because temperature and pressure is constant.
Thus, we can conclude that for this case, temperature and pressure needs to be constant.
So, the correct answer is (B).
Note: Remember that for all spontaneous reactions, the value of dG (Change in Gibbs free energy) is negative (dG < 0) and for all non-spontaneous reactions, dG has a positive value ( dG> 0).
Complete step by step solution:
We know that the state function G is called Gibbs free energy of the system. The value of dG decided whether the reaction will be spontaneous or nonspontaneous.
- Here, we are given the formula of Gibbs free energy G which can be written as
G = H – TS ….(1)
Here, H is the enthalpy, T is the absolute temperature of the system and S is the entropy.
- So, now we need to find which quantities will be constant in the spontaneous process involving only PV work.
We can write the equation (1) in its differential form as
dG = dH – TdS
Now, dH is the enthalpy change and it is related with internal energy (U) and work (W) as
dH = dU + dW ….(2)
Now, we are given that work is of PV type only. So, we can write that dW = PdV. So, we can write the equation (2) as
dH = dU + pdV …(3)
Now, put equation (3) in equation (1), so we will get
dG = dU + PdV – TdS
Now, for a spontaneous process, dG < 0 and TdS > dU+PdV and VdP=SdT=0 because temperature and pressure is constant.
Thus, we can conclude that for this case, temperature and pressure needs to be constant.
So, the correct answer is (B).
Note: Remember that for all spontaneous reactions, the value of dG (Change in Gibbs free energy) is negative (dG < 0) and for all non-spontaneous reactions, dG has a positive value ( dG> 0).
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