Question

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?

Answer in - ATM

Answer 1

Answer:

601.8 mL

Explanation:

For an ideal gas kept at constant temperature and pressure, the number of moles of the gas is directly proportional to the volume of the gas:

$n\propto V$

where

n is the number of moles

V is the volume of the gas

Therefore, for a gas undergoing a transformation at constant temperature and pressure, we can write:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

where here we have:

$V_1=343.4 mL$ is the initial volume of the gas

$n_1=0.214 mol$ is the initial number of moles

$n_2=0.375 mol$ is the final number of moles

Solving for V2, we find the final volume of the gas:

$V_2=\frac{n_2 V_1}{n_1}=\frac{(0.375)(343.4)}{0.214}=601.8 mL$

Answer 2

The volume would 0.375 mol of argon gas occupy under the same conditions will be "601.8 mL".

Pressure and Temperature

According to the question,

Gas initial volume, V₁ = 343.4 mL

Initial no. of moles, n₁ = 0.214 mol

Final no. of moles, n₂ = 0.375 mol

We know that,

Moles of gas ∝ Volume of gas

or,

                   n ∝ V

and, the relation will be:

→ $\frac{V_1}{n_1} = \frac{V_2}{n_2}$

then,

The final volume be:

→ V₂ = $\frac{n_2 V_1}{n_1}$

By substituting the values, we get

       = $\frac{0.375\times 343.4}{0.214}$

       = $\frac{128.775}{0.214}$

       = 601.8 mL  

Thus the above answer is correct.

Find out more information about volume here:

https://brainly.com/question/25290815