# Question 34f5b

Dec 14, 2014

The gas would reach 12 atm pressure at a temperature of 1200 K or 1000 °C.

This is an Ideal Gas Law problem: $P V = n R T$

$P$ = 12 atm
$V$ = 0.525 L
$n$ = 0.062 mol
$R$ = 0.082 06 L·atm·K⁻¹mol⁻¹

So T = (PV)/(nR) ="12 atm × 0.525 L"/("0.062 mol × 0.082 06 L·atm·K"^-1"mol"^-1)# =

1240 K (2 significant figures + 1 guard digit) = 1200 K (2 significant figures)

1240 K = (1240 -273) °C = 1000 °C

Note: The answer has only one significant figure.

The 1240 was really 12.₄ × 10². When we subtract 2.73 ×10², we get 9. ₇ × 10²
(1 significant figure).

To show 1 significant figure, we write this as 10 × 10³ or 1000.