Question

A 2 L sample of N2(g) and a 1 L sample of Ar(g), each originally at 1 atm and 0°C, are combined in a 1 L tank. If the temperature is held constant, what is the total pressure of the gases in the tank? a. 1 atm b. 2 atm c. 3 atm d. 4 atm e. 5 atm

Answer 1

Answer:

c. 3 atm

Explanation:

First, we will use the ideal gas equation to find the moles of each gas.

N₂(g)

P × V = n × R × T

n = P × V / R × T

n = 1 atm × 2 L / (0.082 atm.L/mol.K) × 273 K

n = 0.089 mol

Ar(g)

P × V = n × R × T

n = P × V / R × T

n = 1 atm × 1 L / (0.082 atm.L/mol.K) × 273 K

n = 0.045 mol

The total gaseous moles are 0.089 mol + 0.045 mol = 0.134 mol

Now, we will use the ideal gas equation to find the pressure of the gaseous mixture.

P × V = n × R × T

P = n × R × T / V

P = 0.134 mol × (0.082 atm.L/mol.K) × 273 K / 1 L

P = 3 atm

Answer 2

Answer:

The total pressure in the tank is 3.0 atm

Explanation:

Step 1: Data given

Volume of the N2 sample = 2.0 L

Volume of the Ar sample = 1.0 L

Temperature = 0°C

Volume of the tank = 1L

Step 2: Calculate moles of N2

p*V=n*R*T

n =(p*V)/(R*T)

⇒ with p = the pressure of N2 gas = 1.0 atm

⇒ with V = the volume of N2 = 2.0 L

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

⇒ with T = the temperature = 273.15 K

n = (1.0*2.0)/(0.08206*273.15)

n = 0.0892 moles

Step 3: Calculate moles of Ar

n =(p*V)/(R*T)

⇒ with p = the pressure of Ar gas = 1.0 atm

⇒ with V = the volume of Ar = 1.0 L

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

⇒ with T = the temperature = 273.15 K

n = (1.0*1.0)/(0.08206*273.15)

n = 0.0446 moles

Step 4: Calculate total number of moles

Tot moles = 0.0892 + 0.0446 = 0.1338 moles

Step 5: Calculate total pressure

p*V=n*R*T

p = (n*R*T)/V

⇒ with p = the total pressure  = ?

⇒ with n = the number of moles = 0.1338 moles

⇒ with V = the volume in the tank = 1.0 L

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

⇒ with T = the temperature = 273.15 K

p = (0.1338 * 0.08206 * 273.15)/1

p = 3.0 atm

The total pressure in the tank is 3.0 atm