Chapter 5 - Chemistry Notes II

# Chapter 5 - Chemistry Notes II - -Then the rest can be...

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Gas -Characterized by Pressure (P), Temperature (T(K), Volume and number of moles (M) -Combined gas Law P 1 V 1 /N 1 T 1 = P 2 V 2 /n 2 T 2 T must be in Kelvin (K = ºC + 273) -Ideal gas equation: PV = nRT R = 0.08206 L Atm / mol K Application: Molar mass determination Density -Reaction Stoichiometry: An airbag inflates: 2NaNs (s) 2Na (s) + 3 N 2 (g) ? g of NaN3 must be used to inflate the airbag to 26L Atm pressure of 1.15atm at 25ºC. (Combined) An inflated balloon has a volume of 0.5L at sea level (1 Atm) and is allowed to ascend in altitude until the pressure is 0.45 Atm. During ascent, the temperature of the gas falls from 22ºC to -21ºC. Calculate the volume of the ballon at its final altitude Ideal Gas:? Volume occupied by 16.0 g of ethane at 720 Torr and18ºC? Molar Mass:? Molar mass of a gas at 100ºC if it occupies 185ml with a pressure of 755 torr and its mass is 0.523g.

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Unformatted text preview: -Then the rest can be derived from it-If n, T stay constant it will cancel out P 1 V 1 = P 2 V 2 (Boyle’s Law)-(T 1 = V 2 /T 2 Charles’ Law-P 1 T = cte V 1 /n 1 = V 2 n 2 Avogadro’s Law-Combined Gas Law E.G. Set problem as Initial Final P 1 = 1 Atm P2= 0.45Atm V 1 = 0.5L V2=? T 1 = 22ºC + 273 T2=-21ºC + 273 N 1 = N2= *Usually this is constant T2/P2 P1V1/T1 = P2V2/T2 (T2/P2) V2 = v1 (P1.P2) (T2/T1) = 0.5L (1atm/0.45atm) ( 252K/295K) Ideal Gas Equation: P V = n R T Atm L mol .08206 ( [L(Atm)]/mol(k)]) K Let’s link molar Mass, density to PV = nRT if d = m/v then = m/d d P (m/d ) = n R T x d d = mP/nRT D=MP/RT M is molar mass, g/mol ? molar mass to PV = n R T (g/mol)nM = m/n x n PV – m/M RT n = m/M M = mRT/PV Stoichiometry involving gas E.G. Airbag Rx Mass(g)...
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Chapter 5 - Chemistry Notes II - -Then the rest can be...

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