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Answer 1 · 2015-04-14T17:55:26

The volume is 40.6 L.

The temperature remains constant while $P$ $P$ and $V$ $V$ change.

This is a Boyle's Law problem.

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$

$P_{1} = 2.00 atm$ $P_1 = "2.00 atm"$ ; $V_{1} = 20.3 L$ $V_1 = "20.3 L"$
$P_{2} = 1.00 atm$ $P_2 = "1.00 atm"$ ; $V_{2} = ?$ $V_2 = "?"$

$V_2 = "20.3 L" × (2.00 cancel("atm"))/(1.00 cancel("atm")) = "40.6 L"$

The volume of the gas is 40.6 L