When 1 gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphre, the volume increases from 1cc to 1671cc. The heat of vaporisation at his pressure is 540 cal/gm. Increase in internal energy of water is (1 atmosphre =1.01x `10^(6)` dyne/`cm^(2)`)

Options

A. 4200J
B. 8200J
C. 1200J
D. 2100J

Explanation

To find the increase in internal energy of water when it changes from liquid to vapor at constant pressure, we can use the first law of thermodynamics, which states:

\[ \Delta U = Q - W \]

where:
- \(\Delta U\) = change in internal energy
- \(Q\) = heat added to the system
- \(W\) = work done by the system

### Step 1: Calculate the heat added (Q)

The heat added during the phase change (vaporization) can be calculated using the formula:

\[ Q = m \cdot L \]

where:
- \(m\) = mass of water = 1 gm
- \(L\) = latent heat of vaporization = 540 cal/gm

Substituting the values:

\[ Q = 1 \, \text{gm} \cdot 540 \, \text{cal/gm} = 540 \, \text{cal} \]

### Step 2: Convert heat from calories to joules

To convert calories to joules, we use the conversion factor \(1 \, \text{cal} = 4.2 \, \text{J}\):

\[ Q = 540 \, \text{cal} \cdot 4.2 \, \text{J/cal} = 2268 \, \text{J} \]

### Step 3: Calculate the work done (W)

The work done by the system during the expansion can be calculated using the formula:

\[ W = P \cdot \Delta V \]

where:
- \(P\) = pressure = 1 atm = \(1.01 \times 10^6 \, \text{dyne/cm}^2\)
- \(\Delta V\) = change in volume = \(V_2 - V_1 = 1671 \, \text{cc} - 1 \, \text{cc} = 1670 \, \text{cc}\)

Now substituting the values:

\[ W = 1.01 \times 10^6 \, \text{dyne/cm}^2 \cdot 1670 \, \text{cm}^3 \]

Calculating the work done:

\[ W = 1.01 \times 10^6 \cdot 1670 = 1683670000 \, \text{dyne} \cdot \text{cm}^2 \]

To convert this into joules, we use the conversion \(1 \, \text{J} = 10^7 \, \text{dyne} \cdot \text{cm}^2\):

\[ W = \frac{1683670000}{10^7} = 168.367 \, \text{J} \]

### Step 4: Calculate the change in internal energy (\(\Delta U\))

Now we can find the change in internal energy using the first law of thermodynamics:

\[ \Delta U = Q - W \]

Substituting the values we calculated:

\[ \Delta U = 2268 \, \text{J} - 168.367 \, \text{J} = 2099.633 \, \text{J} \]

Rounding this to two significant figures gives:

\[ \Delta U \approx 2100 \, \text{J} \]

### Final Answer:

The increase in internal energy of water is approximately **2100 J**.

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