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If the atmospheric conditions are 20°C, 1.013 bar and specific humidity of 0.0095 kg/kg of dry air, the partial pressure of vapour will be nearly
1. 0.076 bar
2. 0.056 bar
3. 0.036 bar
4. 0.016 bar

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Correct Answer - Option 4 : 0.016 bar

Concept:

The specific humidity is given by

\(\omega = \frac{{0.622{P_v}}}{{P - {P_v}}}\) 

Where ω is specific humidity, P is pressure of air, Pv is partial pressure of vapour

Calculation:

Given P = 1.013 bar, ω = 0.0095 kg/kg of dry air;

From specific humidity formula,

\(0.0095 = \frac{{0.622{P_v}}}{{1.013 - {P_v}}} \Rightarrow {P_v} = 0.01524\;bar\) 

Nearest option is 0.016 bar

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