I attempted this problem but for some reason i keep getting the wrong answer, to attain the mass i used the equation PV=nRT, solving for n being the number of moles i then multiplied it by the mass per mole given in the question, coming to a conclusion of 365 kg but my instructor listed the correct answer being 357 kg can someone point out what i did wrong so i don't make the same mistake in the future.
What is the mass of the air, to the nearest kilogram, in a room 10 m x 10m x 3 m, if the air temperature is 20 degrees Celsius and atmospheric pressure is 1 x 105 N/m2? (One mol of air has a mass of 0.029 kg)
What is the mass of the air, to the nearest kilogram, in a room 10 m x 10m x 3 m, if the air temperature is 20 degrees Celsius and atmospheric pressure is 1 x 105 N/m2? (One mol of air has a mass of 0.029 kg)
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PV = nRT we want to solve for n, the number of mols, so:
PV/RT = n
now since there's .029kg in ever mole to get the mass in kg we multiply by .029
(.029)PV/RT = mass
(.029)(10^5)(300)/((293)(8.3145)) = 357 kg
PV/RT = n
now since there's .029kg in ever mole to get the mass in kg we multiply by .029
(.029)PV/RT = mass
(.029)(10^5)(300)/((293)(8.3145)) = 357 kg