   Chapter 13, Problem 64QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
110 views

# The “Chemistry in Focus." segment Snacks Need Chemistry, Too! discusses why popcorn “pops." You can estimate the pressure inside a kernel of popcorn at the time of popping by using the Meat gas law. Basically, you determine the mass of water released when the popcorn pups by measuring the mass of the popcorn both before and after popping. Assume that the difference in mass is the mass of water vapor lost on popping. Assume that the popcorn pops at the temperature of the cooking oil ( 225  ° C ) and that the volume at" the “container" is the volume of the unpopped kernel. Although we are making several assumptions, we can at least get some idea of the magnitude 0f the pressure inside the kernel.suming a total volume of 2 .0  mL for 2 0 kernels and a mass of 0. 25 0  g of water lost from them an popping, calculate the pressure inside the kernels just before they “pop."

Interpretation Introduction

Interpretation:

The pressure inside the Kernels should be calculated.

Concept Introduction:

The ideal gas law was purposed to explain all the properties of any ideal gas. The expression for ideal gas law can be shown as:

PV = n R T

Here:

• P = pressure of gas.
• V = volume of gas.
• n = number of moles of gas.
• R = gas constant = 0.0821 L atm / K mol.
• T = Temperature.
Explanation

The volume and temperature values are given as follows:

Volume = 2.0 mL = 0.002 L

Temperature = 225 ° C = 225 + 273.15 = 498.15 K

Mass = 0.250 g

Molar mass of water = 18.0 g/mol

Number of moles can be calculated as follows:

n=mM

Putting the values,

n=0.250 g18.0 g/mol=0.014 mol

Now, PV = n R T

Plug the values in the ideal gas equation to calculate the pressure:

×0

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