A 4.44 L container holds 15.4 g of oxygen at 22.55°C. What is the pressure?

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A 4.44 L container holds 15.4 g of oxygen at 22.55°C. What is the pressure?

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Answer 1 · 2017-06-18T20:06:55

We use the Ideal Gas Equation to get........... $P = 2.63 \cdot a t m$ $P=2.63*atm$

Explanation:

$P = \frac{n R T}{V} = \frac{\frac{15.4 \cdot g}{32.00 \cdot g \cdot m o l^{- 1}} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 295.7 \cdot K}{4.44 \cdot L} =$ $P=(nRT)/V=((15.4*g)/(32.00*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx295.7*K)/(4.44*L)=$

$2.63 \cdot a t m$ $2.63*atm$ .

Do the units cancel out to give an answer in $a t m$ $atm$ ? Do they? Don't assume that I haven't made a mistake!

How did I do this? Well for (i) I know that oxygen is a binuclear gas, i.e. $O_{2}$ $O_2$ ; in fact all the elemental gases (save for the Noble Gases) are BINUCLEAR.....and thus I used a molecular mass of $32.00 \cdot g \cdot m o l^{- 1}$ $32.00*g*mol^-1$ for dioxygen gas.......

And (ii) I knew the gas constant was............... $0.0821 \cdot L \cdot a t m \cdot K^{- 1} \cdot m o l^{- 1}$ $0.0821*L*atm*K^-1*mol^-1$ .

And (iii) I knew the relationship between $^{\circ} C$ $""^@C$ and $degrees Kelvin$ $"degrees Kelvin"$ .

These data (well (ii) and (iii)) need not be remembered, as they should be provided (with various units) in a chemistry exam. You still need to know how to use the appropriate gas constant, $R$ $R$ .

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A 4.44 L container holds 15.4 g of oxygen at 22.55°C. What is the pressure?

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