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.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml