A 2.50 L container is filled with 175 g of argon; a. If the pressure is 10 atm, what is the temperature ? b if the temperature is 22K, what is the pressure ?

A. 70 K

B. 3.16 atm

Explanation:

We'll begin by calculating the number of mole in 175 g of Ar. This can be obtained as follow:

Mass of Ar = 175 g

Molar mass of Ar = 40 g/mol

Mole of Ar =?

Mole = mass / molar mass

Mole of Ar = 175 / 40

Mole of Ar = 4.375 moles

A. Determination of the temperature.

Mole of Ar (n) = 4.375 moles

Volume (V) = 2.50 L

Pressure (P) = 10 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =?

PV = nRT

10 × 2.5 = 4.375 × 0.0821 × T

25 = 4.375 × 0.0821 × T

Divide both side by (4.375 × 0.0821)

T = 25 / (4.375 × 0.0821)

T ≈ 70 K

B. Determination of the pressure.

Mole of Ar (n) = 4.375 moles

Volume (V) = 2.50 L

Temperature (T) = 22 K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =?

PV = nRT

P × 2.5 = 4.375 × 0.0821 × 22

Divide both side by 2.5

P = (4.375 × 0.0821 × 22) / 2.5

P = 3.16 atm.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

im pretty sure its b