# Solution a f mg 101325x10 5 n mg 5 2 kgm 101325 x 10

• CountAtomManatee7149
• 61

This preview shows page 4 - 6 out of 61 pages.

Solution: a) F = 1.01325x10 5 N = 5 2 kg•m 1.01325 x 10 s = ( mass ) (9.81 m/s 2 ) mass = 1.03287x10 4 = 1.03x10 4 kg b) 2 3 2 4 2 kg 10 g 10 m 1.03287x10 1 kg 1 cm m     = 1.03287x10 3 g/cm 2 (unrounded) Height = 3 3 2 g 1 mL 1 cm 1.03287x10 22.6 g 1 mL cm  = 45.702 = 45.7 cm Os 5.16 The statement is incomplete with respect to temperature and mass of sample. The correct statement is: At constant temperature and moles of gas, the volume of gas is inversely proportional to the pressure. 5.17 a) Charles’s law: At constant pressure, the volume of a fixed amount of gas is directly proportional to its Kelvin temperature. Variable: volume and temperature; Fixed: pressure and moles b) Avogadro’s law: At fixed temperature and pressure, the volume occupied by a gas is directly proportional to the moles of gas. Variable: volume and moles; Fixed: temperature and pressure c) Amontons’s law: At constant volume, the pressure exerted by a fixed amount of gas is directly proportional to the Kelvin temperature. Variable: pressure and temperature; Fixed: volume and moles mg mg
5-4
5.18 Plan: Examine the ideal gas law; volume and temperature are constant and pressure and moles are variable.
5.19 Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant so 1 1 1 1 2 2 PV n T T 2 2 = P V n . 2 2 2 . 2 2 V 2 .
c) T is fixed and V doubles. n doubles since one mole of reactant gas produces a total of 2 moles of product gas. 2 n . d) P is fixed and V doubles. n is fixed since 2 moles of reactant gas produce 2 moles of product gas.
2 T .
5.20 Plan: Use the relationship 1 1 2 1 1 2 2 = PV P V n T n T or 1 1 2 2 2 1 1 = PV n T V P n T 2 2 . = 1
• • • 