Physical Chemistry: Thermodynamics, Structure, and Change

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

Chapter 1 - Topic 1A - The perfect gas - Exercises - Page 54: 1A.2(a)

Answer

(i) $3.42 \space bar$ (ii) $3.38 \space atm$

Work Step by Step

Boyle's law: $p_iV_i = p_fV_f$ $$V_i - 2.20 \space dm^3 = V_f = 4.65 \space dm^3$$ $$V_i = 4.65 \space dm^3 + 2.20 \space dm^3 = 6.85 \space dm^3$$ $$p_i(6.85 \space dm^3) = (5.04 \space bar)(4.65 \space dm^3)$$ $$p_i = \frac{(5.04 \space bar)(4.65 \space dm^3)}{6.85 \space dm^3} = 3.42 \space bar$$ $$1 \space atm = 1.01325 \space bar$$ $$3.42 \space bar \times \frac{1 \space atm}{1.01325 \space bar} = 3.38 \space atm$$
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