when a 0.134 g popcorn kernel pops, it loses about 13% of its mass. Assuming tha

Question

when a 0.134 g popcorn kernel pops, it loses about 13% of its mass. Assuming that the volume of a typical kernel is 9.5x10 mL and that the mass loss is due to water lost, use the ideal gas law to calculate the pressure of the water vapor in the kernel immediately before it pops at 100°C. Does the resulting calculated pressure account for the popping of the popcorn 5, The volume of a gas at several temperatures is recorded below. On a piece of graph paper, plot the volume versus the temperature. Is Charles' law apparently obeyed? By extrapolating the graph to a volume of 0.00 L, determine the value of absolute zero in "C according to this data. 6. praYolume da Temporature CO Volume L) 2.33 2.21 2.07 1.88 20 100 80 60 40 2.98 2.87 2.68 2.53 -20 -40 oc T (extrapolated to zero volume

Explanation / Answer

Q5

m = 0.134 g of kernel props

13% is lost

a)

V = 9.5*10^-2 mL = 9.5*10^-5 L

mass is water...

then

get pressure of water T = 100°C = 373 K

13% --> 13/100 * 0.134 = 0.01742 g of water

mol of water = mass/MW = 0.01742/18 = 0.000967 mol of water

now

PV = nRT

P = nRT/V = (0.000967)(0.082*373)/(9.5*10^-5) = 311.3332 atm

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