Question

Calculate the pressure, in atmospheres, of 2.0 mol of helium gas in a 10.0 L container at 27 degrees Celsius.

Answer 1

Answer:

4.92atm

Explanation:

From the question, the following were obtained:

V = 10L

T = 27°C = 27 + 273 = 300K

n = 2moles

R = 0.082atm.L/K /mol

P =?

Using the ideal gas equation PV = nRT, we can solve for the pressure as follows:

PV = nRT

P = nRT / V

P = (2x0.082x300) / 10

P = 4.92atm

Answer 2

The pressure, in atmospheres, of the helium gas is 4.93 atm

From the question,

We are to determine the pressure of the helium gas

Using the formula

PV = nRT

Where P is the pressure

V is the volume

n is the number of moles

R is the gas constant

and T is the temperature

Then, we can write that

$P = \frac{nRT}{V}$

From the given information

n = 2.0 mol

V = 10.0 L

T = 27 °C = 27 + 273.15 K = 300.15K

and

R = 0.08206 L atm mol⁻¹ K⁻¹

Putting the parameters into the formula, we get

$P = \frac{2.0 \times 0.08206 \times 300.15}{10.0}$

$P = \frac{49.260618}{10.0}$

∴ P = 4.9260618 atm

P ≅ 4.93 atm

Hence, the pressure, in atmospheres, of the helium gas is 4.93 atm

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