28.5k views
4 votes
A sample of gas contains 6.25 × 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals?

User Vanowm
by
3.2k points

2 Answers

3 votes

Answer:

1st part: B) volume, C) temperature, D) moles.

2nd part: A) Pressure

3rd part: C) P=nRT/V

4th part: 55.9 kPa

Step-by-step explanation:

Correct edge 2021

User Apqu
by
3.2k points
1 vote

Answer: 55.9 kilopascals

Step-by-step explanation:

Given that:

Volume of gas (V) = 500.0mL

[convert 500.0 mL to liters

If 1000 mL = 1 liter

500.0 mL = 500.0/1000 = 0.5 liters]

Temperature of gas (T) = 265°C

Convert 265°C to Kelvin by adding 273

(265°C + 273 = 538K)

Pressure of gas (P) = ?

Number of moles (n) = 6.25 × 10^-3 mol

Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1

To get pressure, apply ideal gas equation

pV = nRT

p x 0.5L = 6.25 × 10^-3 mol x 0.0821 atm L K-1 mol-1 x 538K

0.5L•p = 0.276 atm L

Divide both sides by 0.5L

0.5L•p / 0.5L = 0.276 atm L / 0.5L

p = 0.552 atm

Since pressure is required in kilopascal, convert 0.552 atm to kilopascal

If 1 atm = 101.325 kPa

0..552 atm = Z

cross multiply

Z x 1 atm = 0.552 atm x 101.325 kPa

Z x 1 atm = 55.9 atm•kPa

Z = 55.9 atm•kPa / 1 atm

Z = 55.9 kPa

Thus, the pressure of the gas in kilopascals is 55.9 kPa

User Alexis Diel
by
3.4k points