Question

A student collects 629ml of oxygen at 0.500at, the student collected 0.0337 moles. At what temperature did the student collect the oxygen?

Answer 1

Answer:

The temperature is 114 K or -159ºC

Explanation:

We must use the ideal gas equation to calculate the temperature in K:

P x V = n x R x T

⇒ T = (P x V)/(n x R)

We have to introduce the data expressed in the adequate units:

V = 629 ml x 1 L/1000 ml= 0.629 L

n = 0.0337 moles

P = 0.500 atm

R = 0.082 L.atm/K.mol (it is the gas constant)

T = (P x V)/(n x R) = (0.500 atm x 0.629 L)/(0.0337 mol x 0.082 L.atm/K.mol)

T = 113.8 K ≅ 114 K

Thus, the temperature is 114 K or -159ºC.

Answer 2

Answer:

114 K

Explanation:

Given data

• Volume of oxygen (V): 629 mL = 0.629 L

• Pressure of oxygen (P): 0.500 atm

• Moles of oxygen (n): 0.0337 mol

• Temperature (T): ?

We can calculate the temperature at which the student collected the oxygen using the ideal gas equation.

$P \times V = n \times R \times T\\T = \frac{P \times V}{n \times R} = \frac{0.500atm \times 0.629L}{0.0337mol \times 0.0821atm.L/mol.K} = 114 K$

The oxygen gas was collected at 114 K.