39.9g oxygen gas are held in a 233.7mL container at 158 o C. What would the temperature need to be

changed to in order for the same gas to be at a pressure of 498kPa and a volume of 388.6mL?

-254°C will be the needed temperature

Explanation:

We use the Ideal Gases Law to solve this problem:

Ideal Gases Law → P. V = n. R. T

If we apply the formula for the two situations, we cancel the n because the number of moles stays the same and we also cancel the R, because it is the same value for both situations (Ideal Gases Constant), so now we have these formulas

P₁. V₁ / T₁ = P₂. V₂ / T₂

We do some conversions to have homogeneous units:

233.7 mL. 1L / 1000 mL = 0.2337 L

386.3 mL. 1L / 1000 mL = 0.3863 L

498 kPa. 1atm / 101.3 kPa = 4.91 atm

First of all, we need the pressure for the first situation:

We convert the mass to moles → 39.9 g / 32g/mol = 1.25 moles

We convert the T°C to T° K → 158°C + 273 = 431K

P. V = n. R. T → P = (n. R. T) / V We replace

P = (1.25 mol. 0.082. 431K) / 0.2337L = 189 atm

With all data, we can replace in the main formula

(189 atm. 0.2337L) / 431 K = (4.91 atm. 0.3886L) / T₂

0.1025 atm. L / K = 1.908 atm. L / T₂

T₂ = 1.908 atm. L / 0.1025 K / atm. L = 18.6 K

We convert the value to T°C → 18.6 K - 273 = - 254°C