Question

a quantity of n2 gas originally held at 4.75 atm pressure in a 1.00-L container at 26c is transerred to a 10.0-L container at 20c. a quanity of o2 gas originally at 5.25 atm and 26c in a 5.00-L container is transferred to this same container. what is the total pressure in the new container

Answer 1

Answer:

The total pressure in the new container is 5.608 atm

Explanation:

Step 1: we determine number of moles of N₂ gas present

PV = nRT

Given;

Pressure of N₂ gas = 4.75 atm

Volume of N₂ gas = 1 L

Temperature N₂ gas = 26 ⁰C

R is constant = 0.0821 atm.L/K.mol

n is the number of moles of N₂ gas = ?

$n = \frac{PV}{RT} = \frac{4.75*1}{0.0821*(26+273)} = 0.1935 \ moles \ of \ N_2$

Step 2: we determine number of moles of O₂ gas present

Given;

Pressure of Given;

Pressure of  O₂ gas = 5.25 atm

Volume of  O₂ gas = 5 L

Temperature N₂ gas = 20⁰C

n is the number of moles of  O₂ gas = ?

$n = \frac{PV}{RT} = \frac{5.25*5}{0.0821*(20+273)} = 1.0912 \ moles \ of \ O_2$

Step 3: we determine the total pressure of gases in the new container.

Total number of moles of gases = 0.1935 + 1.0912 = 1.2847

Total volume of gases = 1 + 5 = 6L

Total temperature = 26 + 20 = 46 ⁰C

$P_{Total} = \frac{nRT}{V} = \frac{1.2847*0.0821*(46+273)}{6} = 5.608 \ atm$

Therefore, the total pressure in the new container is 5.608 atm

Answer 2

Answer:

The pressure in the new container is 3.03atm. The pressure is reduced due to a change in volume from smaller containers to a larger container and also due to a reduction in temperature of the containing vessel.

Explanation:

This problem involves the concept of partial pressure of both gases given in the question. In order to find the pressure in the new container assuming ideal gas behavior, we must first calculate the number of moles of each gas in their original container.

From ideal gas law,

PV = nRT

n = PV/RT

For the N2 gas:

P = 4.75atm, V = 1.00L, R = 0.08206atm•L/mol•K, T = 26°C = (26 + 273) = 299K

n = 4.75×1/0.08206× 299

n = 0.194moles

For the O2 gas:

P = 5.25atm, V = 5.00L, R = 0.08206atm•L/mol•K, T = 26°C = (26 + 273) = 299K

n = 5.25×5.00/0.08206× 299

n = 1.07moles

Total number of moles = 0.194 +1.07 = 1.26moles

P = nRT/V

The new container has volume V = 10.0L and temperature T = 20°C = (20 + 273)K = 293K

Therefore,

P = 1.26×0.08206 × 293/10.0 = 3.03atm

the pressure in the new container is 3.03atm.