Select the correct enthalpy at corresponding temperature using following datas
(i) Heat capacity of solid from 0K to normal melting point 200K
`C_(p.m)(s)=0.035T" "JK^(-1)"mol"^(-1)`
`(ii)` Enthalpy of fusion `=7.5 KJ"mol"^(-1)`
`(iii)` Enthalpy of vaporisation `=30KJ "mol"^(-1)`
`(iv)` Heat capacity of liquid form 200K to normal boiling point 300K
`C_(p.m)(l)=60+0.016T " "JK^(-1)"mol"^(-1)`
`(v)` Heat capacity of gas from 300 K to 600 K at 1 atm
`C_(p.m)(g)=50.0" " JK^(-1)"mol"^(-1)`

Options

A. `S_(200(s))`=7
B. `S_(300(l))`=70.43
C. `S_(300(g))=170.43`
D. `S_(600(g)) = 205.09`

Explanation

To solve the problem of calculating the enthalpy at corresponding temperatures using the provided data, we will follow these steps:

### Step 1: Calculate the Entropy at 200 K (S200)
We need to calculate the entropy of the solid from 0 K to the melting point (200 K) using the given heat capacity function:

\[ C_{p,m}(s) = 0.035T \, \text{JK}^{-1}\text{mol}^{-1} \]

The entropy change can be calculated using the formula:

\[ S_{200} = \int_{0}^{200} C_{p,m}(s) \, dT \]

Substituting the heat capacity:

\[ S_{200} = \int_{0}^{200} 0.035T \, dT \]

Calculating the integral:

\[ S_{200} = 0.035 \cdot \left[ \frac{T^2}{2} \right]_{0}^{200} = 0.035 \cdot \frac{200^2}{2} = 0.035 \cdot 20000 = 700 \, \text{J/mol K} \]

### Step 2: Calculate the Entropy at 200 K Including Fusion (S200F)
Now, we need to include the enthalpy of fusion to find the total entropy at the melting point:

\[ \Delta H_{fusion} = 7.5 \, \text{kJ/mol} = 7500 \, \text{J/mol} \]

The total entropy at 200 K after fusion is:

\[ S_{200F} = S_{200} + \frac{\Delta H_{fusion}}{T_{melt}} = 700 + \frac{7500}{200} = 700 + 37.5 = 737.5 \, \text{J/mol K} \]

### Step 3: Calculate the Entropy at 300 K (S300)
Next, we calculate the entropy of the liquid from 200 K to 300 K using the heat capacity function:

\[ C_{p,m}(l) = 60 + 0.016T \, \text{JK}^{-1}\text{mol}^{-1} \]

The entropy change is:

\[ S_{300} = S_{200F} + \int_{200}^{300} C_{p,m}(l) \, dT \]

Calculating the integral:

\[ S_{300} = 737.5 + \int_{200}^{300} (60 + 0.016T) \, dT \]

Calculating the integral:

\[ \int_{200}^{300} (60 + 0.016T) \, dT = \left[ 60T + 0.008T^2 \right]_{200}^{300} \]

Calculating the values:

\[ = (60 \cdot 300 + 0.008 \cdot 300^2) - (60 \cdot 200 + 0.008 \cdot 200^2) \]
\[ = (18000 + 720) - (12000 + 320) = 18000 + 720 - 12000 - 320 = 6000 + 400 = 6400 \, \text{J/mol K} \]

Thus,

\[ S_{300} = 737.5 + 6400 = 7137.5 \, \text{J/mol K} \]

### Step 4: Calculate the Entropy at 300 K Including Vaporization (S300G)
Now, we move to the gas phase at 300 K. The enthalpy of vaporization is given as:

\[ \Delta H_{vaporization} = 30 \, \text{kJ/mol} = 30000 \, \text{J/mol} \]

The total entropy at 300 K after vaporization is:

\[ S_{300G} = S_{300} + \frac{\Delta H_{vaporization}}{T_{boil}} = 7137.5 + \frac{30000}{300} = 7137.5 + 100 = 7237.5 \, \text{J/mol K} \]

### Step 5: Calculate the Entropy at 600 K (S600G)
Finally, we calculate the entropy of the gas from 300 K to 600 K using the constant heat capacity:

\[ C_{p,m}(g) = 50.0 \, \text{JK}^{-1}\text{mol}^{-1} \]

The entropy change is:

\[ S_{600G} = S_{300G} + \int_{300}^{600} C_{p,m}(g) \, dT \]

Calculating the integral:

\[ S_{600G} = 7237.5 + \int_{300}^{600} 50.0 \, dT = 7237.5 + 50.0 \cdot (600 - 300) = 7237.5 + 50.0 \cdot 300 = 7237.5 + 15000 = 22237.5 \, \text{J/mol K} \]

### Summary of Results
- \( S_{200F} = 737.5 \, \text{J/mol K} \)
- \( S_{300} = 7137.5 \, \text{J/mol K} \)
- \( S_{300G} = 7237.5 \, \text{J/mol K} \)
- \( S_{600G} = 22237.5 \, \text{J/mol K} \)