Find the volume in milliliters of 2.00 mol of an ideal gas at 36°C and a pressure of 1120 torr. Use the ideal gas law and the appropriate Rvalue:
R=0.0821 atmL/mol. K
R= 8.31 kPa • L/mol •K
R= 62.4 torr • L/mol · K
mL

V = 34430 mL

Explanation:

Given data:

Volume in mL = ?

Number of moles of gas = 2.00 mol

Temperature = 36°C (36+273= 309K)

Pressure of gas = 1120 torr

Solution:

Formula:

PV = nRT

V = nRT/P

V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr

V = 38563.2 torr • L / 1120 torr

V = 34.43 L

L to mL

34.43 L ×1000 mL / 1 L

34430 mL