Physical Chemistry: Thermodynamics, Structure, and Change

Published by W. H. Freeman
ISBN 10: 1429290196
ISBN 13: 978-1-42929-019-7

Foundations - Topic A - Matter - Exercises - Page 23: A.15(a)

Answer

(i) $1.47 \times 10^5 \space Pa$ (ii) $1.47 \space bar$

Work Step by Step

1. Identify the conversion factors: You can find these values on Table A.1, in the Resource Section. atm-pascal: $1 \space atm = 101.325 \space kPa$ pascal-bar: $10^{5} \space Pa = 1 \space bar$ kilo($k$) = $10^3$ 2. Convert 1.45 atm into pascal: $$1.45 \space atm \times \frac{101.325 \space kPa}{1 \space atm} = 1.47 \times 10^2 \space kPa$$ $$1.47 \times 10^2 (10^3) \space Pa = 1.47 \times 10^5 \space Pa$$ 3. Calculate the same value in bar: $$1.47 \times 10^5 \space Pa \times \frac{1 \space bar}{10^5 \space Pa} = 1.47 \space bar$$
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