An unknown gas at 130 celsius has a rms speed that is 0.50 that of the rms speed of N2 molecules at 50 degrees celsius. What is the molecular weight of unknown gas?

M = 613.8455 kg/mol

Explanation:

RMS velocity or root mean square velocity has a formula for particles:

RMS = (3*R*T/ M)^ (1/2)

Where R is the ideal gas constant = 8.314 (kg*m^2 / s^2)/ K*mol

T is the temperature in Kelvin

M is the mass of a mole of the gas in kilograms (kg)

1) Convert the temperature to Kelvin for the N2 particles

T = *C +273  

T = 50 + 273

T = 323 K

2) Find the molar mass of the particles in kg for N2

using the periodic table: Nitrogen (N) molar mass is 14.00667 g/mol

N2 molar mass = 2 * 14.00667 = 28.01334 g/mol

Convert to kg/mol using dimensional analysis and conversion factor of 1 kg = 1000 g

28.01334 g / mol * 1 kg / 1000 g = 0.02801334 kg/mol

4) find the RMS of N2

RMS = (3 * (8.314 (kg*m^2 / s^2)/K*mol) * (323K) / 0.02801334 kg/mol))^1/2

=536.2712 m/s

5)If we take 0.5 of N2 RMS or 1/2 we get: 268.1356 m/s for the RMS of the unknown gas

6) Rearrange the equation to solve for M

RMS=(3RT/M)^1/2

Solve for M

We know that a^(1/2) = sqrt(a)

sqrt(RMS) = (3RT/M)

sqrt(RMS) * M = 3RT

M = 3RT / sqrt(RMS)

5) Change the temperature to Kelvins

T = 130 +273 = 403 K

6) Plug the numbers in and solve for M

M = (3*(8.314 (kg*m^2 / s^2)/K*mol) * (403 K)) / sqrt(268.1356 m/s)

M = 613.8455 kg/mol

dynamic equilibrium is the answer.

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'