Question #caa3e

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Question #caa3e

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Answer 1 · 2017-04-27T17:53:33

All of these problems depend on the choice of gas constant $R$ $R$ .....

Explanation:

And typically, chemists, measure volume in $litres$ $"litres"$ , and pressure in $m m \cdot H g$ $mm*Hg$ , where $760 mm Hg \equiv 1 \cdot a t m$ $"760 mm Hg"-=1*atm$ . (NB, you do not use a mercury column to measure pressures > $1 \cdot a t m$ $1*atm$ .

And thus (at least in my opinion) the gas constant that is generally most useful is $R = 0.0821 \cdot L \cdot a t m \cdot K^{- 1} \cdot m o l^{- 1}$ $R=0.0821*L*atm*K^-1*mol^-1$ .

The old standby, $P = \frac{n R T}{V}$ $P=(nRT)/V$ . Which equation is this?

$= \frac{\frac{25 \cdot g}{44.01 \cdot g \cdot m o l^{- 1}} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 298 \cdot K}{0.500 \cdot L} = 27.8 \cdot a t m$ $=((25*g)/(44.01*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx298*K)/(0.500*L)=27.8*atm$ (which is rather high pressure!).

We might have expected the high pressure, because the molar volume at $1 \cdot a t m$ $1*atm$ is approx. $25 \cdot L$ $25*L$ , and here we have compressed the gas into a much smaller volume.

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