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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

Respuesta :

skyluke89

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:

[tex]n\propto V[/tex]

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:

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

where here we have:

[tex]V_1=343.4 mL[/tex] is the initial volume of the gas

[tex]n_1=0.214 mol[/tex] is the initial number of moles

[tex]n_2=0.375 mol[/tex] is the final number of moles

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

[tex]V_2=\frac{n_2 V_1}{n_1}=\frac{(0.375)(343.4)}{0.214}=601.8 mL[/tex]

Cricetus

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:

→ [tex]\frac{V_1}{n_1} = \frac{V_2}{n_2}[/tex]

then,

The final volume be:

→ V₂ = [tex]\frac{n_2 V_1}{n_1}[/tex]

By substituting the values, we get

       = [tex]\frac{0.375\times 343.4}{0.214}[/tex]

       = [tex]\frac{128.775}{0.214}[/tex]

       = 601.8 mL  

Thus the above answer is correct.

Find out more information about volume here:

https://brainly.com/question/25290815

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