Answer : The process is not spontaneous.
Explanation :
As, we know that:
Change in entropy = Change in entropy of system + Change in entropy of surrounding
As we are given in question, the entropy of surroundings decrease by the same amount as the entropy of the system increases.
For the given reaction to be spontaneous, the total change in entropy should be positive.
Given :
Entropy change of system = +125J/K
Entropy change of surroundings = -125J/K
Total change in entropy = Entropy change of system + Entropy change of surroundings
Total change in entropy = 125 J/K + (-125 J/K)
Total change in entropy = 0
The process is at equilibrium because the entropy change is equal to zero. So, the process is not spontaneous.
<h3>
Answer:</h3>
112.08 mL
<h3>
Explanation:</h3>
From the question we are given;
- Initial volume, V1 = 100.0 mL
- Initial temperature, T1 = 225°C, but K = °C + 273.15
thus, T1 = 498.15 K
- Initial pressure, P1 = 1.80 atm
- Final temperature , T2 = -25°C
= 248.15 K
- Final pressure, P2 = 0.80 atm
We are required to calculate the new volume of the gases;
- According to the combined gas law equation;
Rearranging the formula;
Therefore;
Therefore, the new volume of the gas is 112.08 mL
Entropy Change is calculated by (Energy transferred) / (Temperature in kelvin)
deltaS = Q / T
Q = (mass)(latent heat of fusion)
Q = m(hfusion)
Q = (500g)(333J/g) = 166,500J
T(K) = 32 + 273.15 = 305.15K
deltaS = 166,500J / 305.15K
deltaS = 545.63 J/K
