What volume would be occupied by a $0.1 \cdot m o l$ $0.1mol$ quantity of gas at a temperature of $310 \cdot K$ $310K$ , and a pressure of $700 \cdot m m \cdot H g$ $700mmHg$ ?

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What volume would be occupied by a $0.1 \cdot m o l$ $0.1mol$ quantity of gas at a temperature of $310 \cdot K$ $310K$ , and a pressure of $700 \cdot m m \cdot H g$ $700mmHg$ ?

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Answer 1 · 2017-07-25T18:33:29

Well, $P V = n R T$ $PV=nRT$ .........and so....... $V ≅ 3 \cdot L$ $V~=3*L$

Explanation:

You must simply KNOW that $1 \cdot a t m$ $1*atm$ pressure will support a column of mercury that is $760 \cdot m m$ $760*mm$ high; and thus $Torr \equiv 1 \cdot m m \cdot H g$ $"Torr"-=1*mm*Hg$ stands in for a pressure measurement. You must also know the relationship between $degrees Kelvin$ $"degrees Kelvin"$ , and $degrees centigrade$ $"degrees centigrade"$ : $0$ $0$ $^{\circ} C \equiv + 273.15 \cdot K$ $""^@C-=+273.15*K$

$V = \frac{n R T}{P} = \frac{0.1 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 310 \cdot K}{\frac{700 \cdot Torr}{760 \cdot Torr \cdot a t m^{- 1}}}$ $V=(nRT)/P=(0.1*molxx0.0821*(L*atm)/(K*mol)xx310*K)/((700*"Torr")/(760*"Torr"*atm^-1))$

$= ? ? \cdot L$ $=??*L$ . The question would have been better proposed had it been specified that $0.1 \cdot m o l$ $0.1*mol$ quantity of gas had been enclosed in a piston at the given temperature and pressure.

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What volume would be occupied by a $0.1 \cdot m o l$ $0.1mol$ quantity of gas at a temperature of $310 \cdot K$ $310K$ , and a pressure of $700 \cdot m m \cdot H g$ $700mmHg$ ?

1 Answer

Explanation:

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Science

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What volume would be occupied by a 0.1⋅mol0.1*mol quantity of gas at a temperature of 310⋅K310*K, and a pressure of 700⋅mm⋅Hg700*mm*Hg?

1 Answer

Explanation:

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What volume would be occupied by a $0.1 \cdot m o l$ $0.1mol$ quantity of gas at a temperature of $310 \cdot K$ $310K$ , and a pressure of $700 \cdot m m \cdot H g$ $700mmHg$ ?