If 0.214 mol of argon gas occupies a volume of 343.4 mL at a particular temperature and pressure, what volume would 0.375 mol of argon gas occupy under the same conditions?

601.8mL

Explanation:

First, we'll begin by obtaining an equation connecting volume and number of mole together. This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by P

V = nRT/P

Now divide both side by n

V/n = RT/P

Under the same conditions simply means temperature (T), pressure (P) and the gas constant (R) are all constant i. e

RT/P = > constant

Therefore, the above equation can be written as:

V1/n1 = V2/n2

Now we can solve for the final volume otherwise known as the new volume as follow:

Data obtained from the question include:

n1 (initial mole) = 0.214 mol

V1 (initial volume) = 343.4 mL

n2 (final mole) = 0.375 mol

V2 (final volume) = ?

Applying the the equation V1/n1 = V2/n2, the final volume is obtain as illustrated below:

V1/n1 = V2/n2

343.4/0.214 = V2/0.375

Cross multiply to express in linear form

0.214 x V2 = 343.4 x 0.375

Divide both side by 0.214

V2 = (343.4 x 0.375) / 0.214

V2 = 601.8mL

Therefore, 0.375 mol of argon gas will occupy 601.8mL under the same condition.