Question

1 g of water at atmospheric pressure has volume of 1 cc and when boiled it becomes 1681 cc of steam. The heat of vaporisation of water is 540 cal/g. Then the change in its internal energy in this process is

• A. 540 cal
• B. 500 cal
• C. 1681 cal
• D. None of the above

Hint:

Most heat goes to internal energy; some does work against atmosphere.

Answer 1

The Correct Option is B

To determine the change in internal energy when 1 g of water at atmospheric pressure is converted to steam, we can use the formula for the change in internal energy:

The total change in internal energy of a system is given by the equation:

\(\Delta U = Q - W\),

where:

• \(\Delta U\) is the change in internal energy.
• \(Q\) is the heat added to the system.
• \(W\) is the work done by the system.

In this problem, the heat added (\(Q\)) is given as 540 cal (heat of vaporization of water).

The work done (\(W\)) by the system is due to the expansion of the water into steam at atmospheric pressure. Work done at constant pressure is calculated as:

\(W = P \Delta V\),

where:

• \(P\) is the atmospheric pressure, approximately \(1.013 \times 10^5\) Pa in SI units.
• \(\Delta V\) is the change in volume, which is \(1681 \, \text{cc} - 1 \, \text{cc} = 1680 \, \text{cc} = 1.68 \times 10^{-3} \, \text{m}^3\) in SI units.

Convert pressure from Pa to cal/cc, using 1 cal = 4.184 Joules:

1 Pa = \(1 \, \text{N/m}^2 = 1 \, \text{J/m}^3\).

So, W in calories is:

\(W = 1.013 \times 10^5 \times 1.68 \times 10^{-3}\, \text{m}^3 \times \frac{1}{4.184}\, \text{cal/J}\)

Calculating:

\(W \approx 1.013 \times 10^5 \times 1.68 \times 10^{-3} \div 4.184 \approx 40.654\) cal.

Now substitute into the internal energy change formula:

\(\Delta U = Q - W = 540 - 40.654 \approx 499.346\) cal.

Rounding off to the nearest integer, we get \(\Delta U \approx 500\) cal.

Thus, the change in internal energy for this process is approximately 500 cal.

Hence, the correct answer is the option: 500 cal.