If 42.5 mol of an ideal gas is at 8.77 atm and 439 K, what is the volume of the gas?

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If 42.5 mol of an ideal gas is at 8.77 atm and 439 K, what is the volume of the gas?

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Answer 1 · 2018-02-28T08:53:08

Well, is not $P V = n R T$ $PV=nRT$ ?

Explanation:

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

$= \frac{42.5 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 439 \cdot K}{8.77 \cdot a t m}$ $=(42.5*molxx0.0821*(L*atm)/(K*mol)xx439*K)/(8.77*atm)$

$= ? ? \cdot L$ $=??*L$ ...I make this approx. $175 \cdot L$ $175*L$ ...

Answer 2 · 2018-02-28T08:53:41

The volume of the gas is $175$ $175$ liters.

Explanation:

To solve this problem, we need to use the ideal gas law:

$P V = n R T$ $PV=nRT$

$P$ $P$ is pressure in atmospheres, $V$ $V$ is volume in liters, $n$ $n$ is moles, $R$ $R$ is the universal gas constant, and $T$ $T$ is the temperature in Kelvins.

Plug the known values into the equation and solve for the volume.

$(8.77) (V) = (42.5) (0.08206) (439)$ $(8.77)(V)=(42.5)(0.08206)(439)$

$V = \frac{(42.5) (0.08206) (439)}{8.77} = 175$ $V=((42.5)(0.08206)(439))/(8.77)=175$ liters

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If 42.5 mol of an ideal gas is at 8.77 atm and 439 K, what is the volume of the gas?

2 Answers

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Explanation:

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