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Question

A room of 1.5 litres contains gas with pressure 10^5 Pa.If the gas particle has average speed of 750 m/s,then gas mass that trapped in the room is...gram

A.0.8
B.0.7
C.0.6
D.0.5
E.0.4

Answer 1

Given

V= 1.5 L = 1.5 * 10^-3 m^3

P = 10^5 Pa

R = 8.314 J / (mol * K)

Note: J = Pa * m^3

n = number of moles

N = number of particles of the gas

N_a = avogadro number, the number of particles in 1 mole = 6.022 * 10^23 particles / mol

m = n * M

Solution:

v_rms = sqrt(3*RT / M), where M is molar mass in kg / mol

RT = M*v_rms²/3

PV = nRT = (n*M*v_rms²)/3 = (m*v_rms²)/3

m = 3PV / v_rms² = 0.8 * 10^-3 kg = 0.8 grams

Therefore, the answer is A.