The surface area of the floor of a room is 30m2 and the walls are 3m high- If the room temperature is 27-C and pressure is 1 atm- then find the mass of the air inside the room-Given- Molecular weight of air - 29- 1 atm - 1-013 - 105 Pa-

PV = nRT (1.013 × 10^5)(30 × 3)=no. of molesn× 8.31 ×Kelvin(27+278) n=3.65 × 10^3 moles Mass of air inside = nM = 3.65 × 10^3 × 29 g ...

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Standard XIII

Physics

Ideal Gas Equation from KTG

The surface a...

Question

The surface area of the floor of a room is $30 m^{2}$ and the walls are 3m high. If the room temperature is $27^{\circ}$ C and pressure is 1 atm, then find the mass of the air inside the room
[Given: Molecular weight of air = 29, 1 atm = 1.013 $\times 10^{5}$ Pa]

105.85 kg

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29 kg

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74.37 kg

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135.35 kg

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Solution

The correct option is A 105.85 kg
PV = nRT
$(1.013 \times 10^{5}) (30 \times 3) = n o . o f m o l e s n \times 8.31 \times K e l v i n (27 + 278)$
$n = 3.65 \times 10^{3} m o l e s$
Mass of air inside $= n M = 3.65 \times 10^{3} \times 29 g$

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The surface area of the floor of a room is 30m2 and the walls are 3m high. If the room temperature is 27∘C and pressure is 1 atm, then find the mass of the air inside the room [Given: Molecular weight of air = 29, 1 atm = 1.013 ×105 Pa]

The correct option is A 105.85 kgPV = nRT (1.013×105)(30×3)=no.ofmolesn×8.31×Kelvin(27+278) n=3.65×103 moles Mass of air inside =nM=3.65×103×29 g

The surface area of the floor of a room is $30 m^{2}$ and the walls are 3m high. If the room temperature is $27^{\circ}$ C and pressure is 1 atm, then find the mass of the air inside the room
[Given: Molecular weight of air = 29, 1 atm = 1.013 $\times 10^{5}$ Pa]

The correct option is A 105.85 kg
PV = nRT
$(1.013 \times 10^{5}) (30 \times 3) = n o . o f m o l e s n \times 8.31 \times K e l v i n (27 + 278)$
$n = 3.65 \times 10^{3} m o l e s$
Mass of air inside $= n M = 3.65 \times 10^{3} \times 29 g$