Question

A sample of oxygen at 45 degrees Celsius occupies 839 mL. If this sample later occupies 1032 mL at 58 degrees Celsius and 1.9 atm, what was its original pressure?

Answer in atm

Answer 1

1.3 atm

Step-by-step explanation:

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

(P1 x V1) / (T1) = (P2 x V2) / (T2)

where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, and P2, V2, and T2 are the final pressure, volume, and temperature of the gas.

We are given:

P1 is what we want to find

V1 = 839 mL

T1 = 45 degrees Celsius = 318 K

V2 = 1032 mL

T2 = 58 degrees Celsius = 331 K

P2 = 1.9 atm

Plugging these values into the formula, we get:

(P1 x 839 mL) / (318 K) = (1.9 atm x 1032 mL) / (331 K)

Simplifying and solving for P1, we get:

P1 = (1.9 atm x 1032 mL x 318 K) / (839 mL x 331 K)

P1 = 1.3 atm (rounded to one decimal place)

Therefore, the original pressure of the oxygen sample was 1.3 atm.