If a 3.6 moles of an ideal gas occupies 150 mL tank at a temperature of -5 °C, what is the pressure inside the tank?

Question

I'm struggling some with the formulas with this unit so a walk through would be greatly appreciated.

Answer 1

The Ideal Gas Law has units built into it for the value of R used. PV = nRT. P is pressure in atmospheres; V is volume in liters, n is moles based on grams; T is temperature in degrees Kevin. R = 0.08206.

For this problem: V = 0.150 L; n = 3.6; T = 273 - 5 or 268.

Substitution in: P ( 0.150) = 3.6 ( 0.08206) (268)

Divide both sides by 0.15: P = [3.6(0.08206) (268)]/0.150 =527.8 atmospheres

But one must look at significant figures: The volume has 3 since it is taken out to the hundredths place. The moles have 2 figures.

The temperature looks like 1 but has three since you add it to 273 to convert from Centigrade to Kevin.

Thus the answer must have only 2 significant figures since the lowest given is used.

Rounding 527.8 atmospheres to 2 figures gives 530 atmospheres

But to make sure one knows how many figures are significant one must use scientific notation.

530 becomes 5.3 * 10^2.