720 mL of $H_{2}$ $H_2$ gas at 0 degree C and 126.6 kPa is changed to S .T. P. What will be its new volume?

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Answer 1 · 2017-01-10T17:34:25

The new volume will be 907 mL.

STP is defined as 0 °C and 1 bar (100 kPa).

Since only the pressure changes, we can use Boyle's Law to solve the problem.

Boyle's Law is

$color(blue)(bar(ul(|color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "$

We can rearrange Boyle's Law to get

$V_2 = V_1 × P_1/P_2$

In this problem,

$P_1 = "126 kPa"; V_1 = "720 mL"$
$P_2 ="100 kPa"; V_2 = "?"$

∴ $V_2 = "720 mL" × (126 color(red)(cancel(color(black)("kPa"))))/(100 color(red)(cancel(color(black)("kPa")))) = "907 mL"$

720 mL of H_2H2H_2 gas at 0 degree C and 126.6 kPa is changed to S .T. P. What will be its new volume?