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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

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dsdrajlin

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

c. 3 atm

 

First, we will calculate the number of moles by using the ideal gas equation as follows:-

[tex]PV=nRT[/tex].....(1)

Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is the temperature.

N₂(g):-

Substitute 1 atm for pressure(P), 2L for V, 0.0821 Latm\Kmol for R, 273 K for temperature n the equation (1) as follows:-

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

n = 0.089 mol

Ar(g):-

Substitute 1 atm for pressure(P), 1 L for V, 0.0821 Latm\Kmol for R, 273 K for temperature n the equation (1) as follows:-

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

n = 0.0446 mol

The total gaseous moles are 0.089 mol + 0.0446 mol = 0.1336 mol

Now, we will calculate the total pressure by using equation (1) as follows:-

[tex]P = \frac{n \times R \times T}{V}[/tex]

[tex]P = \frac{0.1336\ mol \times 0.0821\ Latm/Kmol \times 298\ K}{1\ L}\\\\=3.26\ atm[/tex]

So, the total pressure of gases in the tank is 3 atm.

Hence, the correct option is (c).

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