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If 4.50 g of methane gas (CH4) is in a 2.00-L container at 35°C, what is the pressure in the

container?

2 Answers

2 votes

Answer:

The pressure in the container is 3.55 atm

Step-by-step explanation:

Step 1: Data given

Mass of methane gas = 4.50 grams

Volume = 2.00 L

Temperature = 35 °C = 308 K

Molar mass CH4 = 16.04 g/mol

Step 2: Calculate moles CH4

Moles CH4 = mass CH4 / molar mass CH4

Moles CH4 = 4.50 grams / 16.04 g/mol

Moles CH4 = 0.281 moles

Step 3: Calculate pressure

p*V = n*R*T

⇒with p = the pressure in the container = TO BE DETERMINED

⇒with V = the volume of the container = 2.00L

⇒with n = the moles of methane = 0.281 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ T = the temperature = 308 K

p = (n*R*T)/V

p = (0.281 * 0.08206 * 308 ) / 2.00

p = 3.55 atm

The pressure in the container is 3.55 atm

User Ben Hirschberg
by
3.8k points
7 votes

Answer:

The answer to your question is Pressure = 3.55 atm

Step-by-step explanation:

Data

R = 0.082 atm L/mol°K

mass = 4.5 g

volume = 2 L

temperature = 35°C

Pressure = ?

Process

1.- Calculate the moles of Methane

molar weight = 12 + 4 = 16 g

16 g of Methane --------------- 1 mol

4.5 g --------------- x

x = (4.5 x 1)/16

x = 0.281 moles

2.- Convert temperature to Kelvin

°K = 273 + 35

= 308

3.- Use the ideal gas law to solve this problem

PV = nRT

- Solve for P

P = nRT / V

- Substitution

P = (0.281)(0.082)(308) / 2

- Simplification

P = 7.097 / 2

- Result

P = 3.55 atm

User BenTrofatter
by
3.7k points