A perfectly conducting vessel of volume V = $0.4 m^{3}$ contains an ideal gas constant temperature T = 273 K. A portion of the gas is let our and the pressure of the gas falls by $Δ P = 0.24 a t m$ (Density of the gas at STP is $ρ = 1.2 k g / m^{3}$ ). Find the mass of the gas which excapes from the vessels.

115.2 g

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215.2 g

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915.2 g

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415.2 g

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Solution

The correct option is A $115.2 g$
The molar mass of gas be M.

Hence $ρ = \frac{P M}{R T}$

$1.2 = \frac{10^{5} (M)}{(8.314) (273)}$

$M = 27.24 g$

Also, since temperature and volume are constant,

$V Δ P = \frac{Δ m}{M} R T$

$Δ m = \frac{M V Δ P}{R T} = \frac{27.24 (0.4) (0.24 X 10^{5})}{(8.314) (273)} = 115.22 g$

Answer is $115.2 g$

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A perfectly conducting vessel of volume V = 0.4m3 contains an ideal gas constant temperature T = 273 K. A portion of the gas is let our and the pressure of the gas falls by ΔP=0.24atm (Density of the gas at STP is ρ=1.2kg/m3). Find the mass of the gas which excapes from the vessels.

The correct option is A 115.2gThe molar mass of gas be M.Hence ρ=PMRT1.2=105(M)(8.314)(273)M=27.24gAlso, since temperature and volume are constant,VΔP=ΔmMRTΔm=MVΔPRT=27.24(0.4)(0.24X105)(8.314)(273)=115.22gAnswer is 115.2g