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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 11 - Problem 66p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 11 - Problem 66p

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Use the ideal gas law to complete the table.PVnT2.39 atm1.21 L______205 K512

ISBN: 9780321910295 34

Solution for problem 66P Chapter 11

Introductory Chemistry | 5th Edition

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Problem 66P

PROBLEM 66P

Use the ideal gas law to complete the table.

 P V n T 2.39 atm 1.21 L ______ 205 K 512 torr ______ 0.741 mol 298 K 0.433 atm 0.192 L 0.0131 mol ______ ______ 20.2 mL 5.71 × 10−3 mol 20.4 °C

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to complete the given table.

Step1:

Boyle’s law, Charles’s law and Avogadro's law can be combined together to form a single equation which is known as ideal gas equation.

At constant T and n; V ∝ 1 / p        Boyle’s law

At constant p and n; V ∝ T                Charles’ law

At constant p and T; V ∝ n                Avogadro’s law

Thus,

V ∝ nT / p

V = R

Where R is the proportionality constant called Universal gas constant. On rearranging the above equation we get,

pV = nRT        ----(1)

Equation (1) is called ideal gas equation.

Step2:

The value of R is 8.20578 x 10-2 L atm K-1 mol-1

Using this value, let us complete the given table.

 P V n T 2.39 atm 1.21 L 0.1719 mol 205 K 512 torr 26.88 L 0.741 mol 298 K 0.433 atm 0.192 L 0.0131 mol 77.3 K 6.877 atm 20.2 mL 5.71 × 10−3 mol 20.4 °C

1. Given, p = 2.39 atm

V = 1.21 L

T = 205 K

From equation (1), n = pV / RT

= (2.39 atm x 1.21 L) / (8.20578 x 10-2 L atm K-1 mol-1  x 205 K)

= (2.8919 atm L) / (16.82 L atm mol-1)

= 0.1719 mol.

b)        Given, p = 512 torr = 512 x 0.00131579 atm = 0.674 atm

n = 0.741 mol

T = 298 K

From equation (1), V = nRT / p

= (0.741 mol x 8.20578 x 10-2 L atm K-1 mol-1  x 298 K) / (0.674 atm)

= (18.12 L atm) / (0.674 atm)

= 26.88 L.

c)        Given, p = 0.433 atm

V = 0.192 L

n = 0.0131 mol

From equation (1), T = pV / nR

= (0.433 atm x 0.192 L) / (0.0131 mol x 8.20578 x 10-2 L atm K-1 mol-1)

= (0.083136 atm L) / (0.001075 L atm K-1)

= 77.3 K(approx.).

d)        Given, V = 20.2 mL = 20.2 / 1000 L= 0.020 L

n = 5.71 x 10-3 mol

T = 20.4 oC = 20.4 + 273.15 K = 293.55 K

From equation (1), p = nRT / V

= (5.71 x 10-3 mol x 8.20578 x 10-2 L atm K-1 mol-1  x 293.55 K) / (0.020 L)

= (0.13754 L atm) / (0.020 L)

= 6.877 atm.

--------------------

Step 2 of 3

Step 3 of 3

ISBN: 9780321910295

The answer to “Use the ideal gas law to complete the table.PVnT2.39 atm1.21 L______205 K512 torr______0.741 mol298 K0.433 atm0.192 L0.0131 mol____________20.2 mL5.71 × 10?3 mol20.4 °C” is broken down into a number of easy to follow steps, and 23 words. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The full step-by-step solution to problem: 66P from chapter: 11 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. Since the solution to 66P from 11 chapter was answered, more than 1979 students have viewed the full step-by-step answer. This full solution covers the following key subjects: mol, atm, Law, ideal, gas. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295.

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