Question

A 800mL of gas sample at STP is compressed to a volume of 200 mL, and temperature increased to 30 °C, what will be the new pressure in kpa.

Answer 1

Answer:

449.7 kPa

Explanation:

We prepare the Ideal Gases Law to solve this question:

P . V = n . R . T

As moles of gas are the same, and R is constant, we can compare this two situations:

P₁ . V₁ / T₁  and P₂ . V₂ / T₂

As this remains constant we can propose:

P₁ . V₁ / T₁  = P₂ . V₂ / T₂

We convert T° to Absolute value: 30°C + 273 = 303 K

We convert the volume from mL to L

800 mL . 1L /1000 mL = 0.8 L

200 mL . 1L /1000 mL = 0.2 L

1 atm . 0.8 L / 273 K = P₂ . 0.2L / 303k

(0.8 L.atm / 273K ) . 303k = P₂ . 0.2L

0.8879 L.atm / 0.2L = 4.4 atm

We convert pressure from atm to kPa

4.4 atm . 101.3 kPa / 1 atm = 449.7 kPa

Answer 2

Answer:

450. kPa

Explanation:

Use the combined gas law. Solve for the unknown. Plug in the numbers, turn the crank, and out pops the answer. A typical plug-n-chug problem.

P1V1 / T1 = P2V2 / T2 …………….. combined gas law

P2 = P1V1T2 / (T1V2) ……………… solve for P2

P2 = 101.3 kPa x 800. mL x 303K / 273K / 200. mL

P2 = 450. kPa

We are assuming 3 significant digits, simply because it makes more sense than having only one significant digit. You didn’t actually specify a pressure unit, but I’m guessing you mean kilopascal, (kPa).