# .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?

## .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?

Mar 4, 2018

$6.5 a t m$

#### Explanation:

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

So,$P V = n R T$

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

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

We get,$P = 6.5 a t m$

Mar 4, 2018

${P}_{\text{container}} = 6.50 \cdot a t m$

#### Explanation:

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

$= \frac{0.54 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 293.15 \cdot K}{2.00 \cdot L}$

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

$\text{absolute temperature"="degree Celsius + 273.15} \cdot K$

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

$= \frac{0.54 \cdot \cancel{m o l} \times 0.0821 \cdot \frac{\cancel{L} \cdot a t m}{\cancel{K \cdot m o l}} \times 293.15 \cdot \cancel{K}}{2.00 \cdot \cancel{L}}$