Question

Container A holds 787 mL of ideal gas at 2.10 atm . Container B holds 154 mL of ideal gas at 4.80 atm . If the gases are allowed to mix together, what is the partial pressure of each gas in the total volume?

Answer 1

You haven't given me temperature so i'll use 300 Kelvin

Step-by-step explanation:

For Container A:

P_A = 2.10 atm

V_A = 787 mL = 0.787 L

R = 0.0821 L.atm/mol.K

T_A = 300 K

Using the ideal gas law:

n_A = (P_A * V_A) / (R * T_A)

n_A = (2.10 atm * 0.787 L) / (0.0821 L.atm/mol.K * 300 K)

n_A ≈ 0.0657 moles

Now, for Container B:

P_B = 4.80 atm

V_B = 154 mL = 0.154 L

R = 0.0821 L.atm/mol.K

T_B = 300 K

Using the ideal gas law:

n_B = (P_B * V_B) / (R * T_B)

n_B = (4.80 atm * 0.154 L) / (0.0821 L.atm/mol.K * 300 K)

n_B ≈ 0.0298 moles

Now that we have the number of moles for each gas in each container, we can find their partial pressures.

Partial pressure of gas A (P_A):

P_A = (n_A * R * T_A) / V_A

P_A = (0.0657 moles * 0.0821 L.atm/mol.K * 300 K) / 0.787 L

P_A ≈ 2.10 atm

Partial pressure of gas B (P_B):

P_B = (n_B * R * T_B) / V_B

P_B = (0.0298 moles * 0.0821 L.atm/mol.K * 300 K) / 0.154 L

P_B ≈ 4.80 atm

Now you have the partial pressures of each gas in the mixture:

Partial pressure of gas A (P_A) ≈ 2.10 atm

Partial pressure of gas B (P_B) ≈ 4.80 atm