.54 mol of H2 is contained in a 2.00 L container at 20.0 oC. What is the pressure in the container in atm?

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.54 mol of H2 is contained in a 2.00 L container at 20.0 oC. What is the pressure in the container in atm?

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Answer 1 · 2018-03-04T10:50:22

$6.5 a t m$ $6.5 atm$

Explanation:

Using ideal gas law to calculate the pressure of the gas,

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

Given values are, $V = 2 L, n = 0.54 m o \leq, T = (273 + 20) = 293 K$ $V=2L,n=0.54 mole,T=(273+20)=293K$

Using, $R = 0.0821 L$ $R=0.0821 L$ atm mol^-1K^-1

We get, $P = 6.5 a t m$ $P=6.5 atm$

Answer 2 · 2018-03-04T11:20:57

$P_{container} = 6.50 \cdot a t m$ $P_"container"=6.50*atm$

Explanation:

We assume Ideality....and thus... $P = \frac{n R T}{V}$ $P=(nRT)/V$

$= \frac{0.54 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 293.15 \cdot K}{2.00 \cdot L}$ $=(0.54*molxx0.0821*(L*atm)/(K*mol)xx293.15*K)/(2.00*L)$

Clearly, I used .........

$absolute temperature = degree Celsius + 273.15 \cdot K$ $"absolute temperature"="degree Celsius + 273.15"*K$

And the expression gives us an answer with units of pressure, as is required....

$=(0.54*cancel(mol)xx0.0821*(cancelL*atm)/cancel(K*mol)xx293.15*cancelK)/(2.00*cancelL)$

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.54 mol of H2 is contained in a 2.00 L container at 20.0 oC. What is the pressure in the container in atm?