no of moles of N2 = W/G.M.Wt
= 0.75/28 = 0.0268moles
no of moles of CH4 = W/G.M.Wt
= 0.125/16 = 0.0078125moles
total no of moles of N2 and CH4 = nN2 + nCH4
= 0.0268+0.0078125
= 0.0346moles
mole fraction of CH4 = nCH4/nCH4 + nN2
= 0.0078125/0.0346 = 0.226
partial pressure of CH4 = mole fraction of CH4* total pressure
4.5 = 0.226*total pressure
total pressure = 4.5/0.226 = 19.9Kpa
P = 19.9Kpa
= 19.9*0.00986923 [ 1KPa = 0.00986923atm]
=0.1964atm
T = 300K
n = 0.0346moles
PV = nRT
V = nRT/P
= 0.0346*0.0821*300/0.1964
= 4.32L
c. 4.32L >>>answer
A gas sample used to simulate the atmosphere of an exoplanet contains 7.50 x 102 mg...
A gas sample used to simulate the atmosphere of an exoplanet contains 7.50 x 102 mg N2 and 1.25 x 102 mg CH4. The partial pressure of CH4 at 300.0 K is 4.5 kPa. What volume does the entire gas sample occupy? 68.9 L 0.550 L 4.32 L 34.0 L 0.693 L
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