Calculate change in enthalpy when 2 moles of liquid water at 1 bar and `100^(@)` C is coverted into steam at 2 bar and `300^(@)`. Assume `H_(2)O` vapoures to behave ideally.
[Latent heat of vaporisation of `H_(2)O(l)` at 1 bar and `100^(@)` C id=s 10.8 kcal per mole]
[R=`2cal//"mol" K`]

Options

A. 21.6 kcal
B. 11.8 kcal
C. 24.8 kcal
D. 23.6 kcal

Answer

To calculate the change in enthalpy when 2 moles of liquid water at 1 bar and 100°C is converted into steam at 2 bar and 300°C, we will follow these steps:

### Step 1: Calculate the enthalpy change for vaporization (ΔH1)

The latent heat of vaporization of water at 1 bar and 100°C is given as 10.8 kcal per mole. Since we have 2 moles of water, we can calculate ΔH1 as follows:

\[ \Delta H_1 = \text{Latent heat of vaporization} \times \text{Number of moles} \]
\[ \Delta H_1 = 10.8 \, \text{kcal/mol} \times 2 \, \text{moles} = 21.6 \, \text{kcal} \]

### Step 2: Calculate the enthalpy change for heating the steam (ΔH2)

Next, we need to calculate the enthalpy change when the steam is heated from 100°C to 300°C at a pressure of 2 bar. The formula for this is:

\[ \Delta H_2 = N C_p \Delta T \]

Where:
- \(N\) = number of moles = 2
- \(C_p\) = molar heat capacity of water vapor = 2 cal/mol K (given)
- \(\Delta T\) = change in temperature = \(300°C - 100°C = 200°C\)

Now substituting the values:

\[ \Delta H_2 = 2 \, \text{moles} \times 2 \, \text{cal/mol K} \times 200 \, \text{K} \]
\[ \Delta H_2 = 800 \, \text{cal} \]

Since we need this in kcal, we convert it:

\[ \Delta H_2 = \frac{800 \, \text{cal}}{1000} = 0.8 \, \text{kcal} \]

### Step 3: Calculate the total change in enthalpy (ΔH)

Now, we can find the total change in enthalpy by adding ΔH1 and ΔH2:

\[ \Delta H = \Delta H_1 + \Delta H_2 \]
\[ \Delta H = 21.6 \, \text{kcal} + 0.8 \, \text{kcal} = 22.4 \, \text{kcal} \]

### Final Answer

The change in enthalpy when 2 moles of liquid water at 1 bar and 100°C is converted into steam at 2 bar and 300°C is **22.4 kcal**.

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