Question

A room contains 35 kg of dry air and 0.5 kg of water vapour. The total pressure and temperature of the air in the room are 100 kPa and 25°C respectively. Given that the saturation pressure for water at 25°C is 3.17 kPa, the relative humidity of the air in the room is
1. 67%
2. 55%
3. 83%
4. 71%

Answer 1

Correct Answer - Option 4 : 71%

Concept:

Specific humidity (ω) can be calculated by:

\(\omega = \frac{{{m_v}}}{{{m_a}}} = 0.621 \times \frac{{{P_v}}}{{{P_{atm}}~ -~ {P_v}}}\)

where m_v = mass of vapour, m_a = mass of dry air, P_v = vapour pressure, P_atm = atmospheric pressure

Relative humidity (ϕ) can be calculated by:

\(\phi = \frac{{{m_v}}}{{{m_{vs}}}} = \frac{{{P_v}}}{{{P_{vs}}}}\)

where m_vs = mass of saturated vapour, P_vs = saturated vapour pressure

Calculation:

Given:

M_a = 35 kg, m_v = 0.5 kg, temperature (t) = 25°c, P_atm = 100 kPa, P_vs = 3.17 kPa

Specific humidity (ω)

\(\omega = \frac{{{m_v}}}{{{m_a}}} = \frac{{0.5}}{{35}} = \frac{1}{{70}}\)

Specific humidity (ω) can also be calculated by:

\(\omega = 0.621 \times \frac{{{P_v}}}{{{P_{atm}} - {P_v}}}\)

\(\frac{1}{{70}} = 0.621 \times \frac{{{P_v}}}{{100 - {P_v}}}\)

100 – P_v = 43.47 × P_v

44.37 × P_v = 100

\({P_v} = \frac{{100}}{{44.37}} = 2.253\;kPa\)

Relative humidity (ϕ)

\(\phi = \frac{{{P_v}}}{{{P_{vs}}}} = \frac{{2.253}}{{3.17}} = 0.71 = 71\;\% \)