Question

When 0.215 mol of a gas is placed into a 34.25 mL container at 125.0°C, 1

Answer 1

Answer:

205.12 atm

Explanation:

Using the ideal gas law equation:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

R = 0.0821 Latm/perK)

T = temperature (K)

n = number of moles (mol)

According to the information in this question;

P = ?

V = 34.25 mL = 34.25 ÷ 1000 = 0.03425L

n = 0.215 mol

T = 125.0°C = 125 + 273 = 398K

Using PV = nRT

P = nRT ÷ V

P = (0.215 × 0.0821 × 398) ÷ (0.03425)

P = 7.025 ÷ 0.03425

P = 205.12 atm