A room contains 35 kg of dry air and 0-5 kg of water vapor- The total pressure and temperature of 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

M_a = 35 kg; m_v = 0.5 kg ω = m_vm_a = 0.535 = 0.01428 also ω = 0.622 p_vp_a p_t = 100 kPa ∴ω = 0.622 p_vp_t p_v Now, 0.01428 = 0.622 p_v100 p_v p_v = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Avogadro's Law & Gay Lussac's Law

A room contai...

Question

A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and $25^{\circ} C$ respectively. Given that the saturation pressure for water at $25^{\circ} C,$ is 3.17 kPa, the relative humidity of the air in the room is

67%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

55%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

83%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

71%

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 71%
$m_{a} = 35 kg; m_{v} = 0.5 kg$

$ω = \frac{m_{v}}{m_{a}} = \frac{0.5}{35} = 0.01428$

also $ω = 0.622 \frac{p_{v}}{p_{a}}$

$p_{t} = 100 kPa$

$∴ ω = 0.622 \frac{p_{v}}{p_{t} - p_{v}}$

Now,
$0.01428 = 0.622 \frac{p_{v}}{100 - p_{v}}$

$p_{v} = 2.238 kPa$

Relative humidity,

$ϕ = \frac{p_{v}}{p_{v s}} = \frac{2.238}{3.17} \times 100$

$= 70.6 % \approx 71 %$

Suggest Corrections

Similar questions

Q. The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar,

30^{\circ} C

and 70%, respectively. If the saturated steam pressure at

30^{\circ} C

is 4.25 kPa, the specific humidity of the room air in kg of water vapour/kg dry air is

Q. Moist air is filled inside a room with total pressure of 100 kPa, and at

25^{\circ}

C. If the moist air is saturated with mass of dry air as 50 kg, the mass of vapour in the room is (use saturation pressure of water at

25^{\circ} C

as 3.17 kPa)

Q. A moist air sample has dry bulb temperature of

30^{\circ} C

and specific humidity of 11.5 g water vapour per kg dry air. Assume molecular weight of air as 28.93. If the saturation vapour pressure of water at

30^{\circ} C

is 4.24 kPa and the total pressure is 90 kPa, then the relative humidity (in %) of air sample is

The condition of air in a closed room is described as follows. Temperature = $25^{0}$ C, relative humidity = 60 %, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure ? The saturation vapour pressure at $25^{0}$ C = 3.2 kPa.

Q. The condition of air in a closed room is described as follows. Temperature = 25°C, relative humidity = 60%, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure? The saturation vapour pressure at 25°C − 3.2 kPa.

Join BYJU'S Learning Program

A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of 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

The correct option is D 71%ma=35kg;mv=0.5kg ω=mvma=0.535=0.01428 also ω=0.622pvpa pt=100kPa ∴ω=0.622pvpt−pv Now, 0.01428=0.622pv100−pv pv=2.238kPa Relative humidity, ϕ=pvpvs=2.2383.17×100 =70.6%≈71%

A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and $25^{\circ} C$ respectively. Given that the saturation pressure for water at $25^{\circ} C,$ is 3.17 kPa, the relative humidity of the air in the room is