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

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

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 = 4.9260618 atm

P ≅ 4.93 atm

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

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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

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

the answer is the first option

explanation:

just googled it m8. not that hard

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