What volume is occupied by a $2 \cdot m o l$ $2mol$ quantity of oxygen, nitrogen, and fluorine gases, at $300 \cdot K$ $300K$ under a pressure of $1.3 \cdot a t m$ $1.3*atm$ ?

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What volume is occupied by a $2 \cdot m o l$ $2mol$ quantity of oxygen, nitrogen, and fluorine gases, at $300 \cdot K$ $300K$ under a pressure of $1.3 \cdot a t m$ $1.3*atm$ ?

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Answer 1 · 2017-09-08T16:07:58

Approx. $40 \cdot L$ $40*L$ ..........

Explanation:

We use the old ideal gas law.....

$V = \frac{n R T}{P}$ $V=(nRT)/P$

We must simply KNOW that oxygen, and nitrogen, and fluorine, and chlorine, are DIATOMIC gases, i.e. $X_{2}$ $X_2$ ...(this is the next refinement of this problem, they could have mentioned a mass of $64 \cdot g$ $64*g$ ....)

And so.......

$V = \frac{2 \cdot m o l \times 0.0821 \cdot L \cdot a t m \cdot K^{- 1} \cdot m o l^{- 1} \times 300 \cdot K}{1.3 \cdot a t m}$ $V=(2*molxx0.0821*L*atm*K^-1*mol^-1xx300*K)/(1.3*atm)$

$= ? ? L$ $=??L$ ....Do the units in the calculation cancel out to give an answer in litres? They should, and if they don't then somewhere we have made an error.

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What volume is occupied by a $2 \cdot m o l$ $2mol$ quantity of oxygen, nitrogen, and fluorine gases, at $300 \cdot K$ $300K$ under a pressure of $1.3 \cdot a t m$ $1.3*atm$ ?

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Explanation:

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What volume is occupied by a 2⋅mol2*mol quantity of oxygen, nitrogen, and fluorine gases, at 300⋅K300*K under a pressure of 1.3⋅atm1.3*atm?

1 Answer

Explanation:

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What volume is occupied by a $2 \cdot m o l$ $2mol$ quantity of oxygen, nitrogen, and fluorine gases, at $300 \cdot K$ $300K$ under a pressure of $1.3 \cdot a t m$ $1.3*atm$ ?