Hello!
The pressure of an 18 L container which holds 16,00 grams of oxygen gas (O₂) at 45 °C is 0,725 atm
To solve this problem we first need to set up the data in the appropriate units to input it in the Ideal Gas Law.
a) 16 g of Oxygen gas to moles of oxygen gas:
[tex]16gO_2* \frac{1 mol O_2}{32gO_2}=0,5 mol O_2[/tex]
b) 45 °C to K
[tex]K=$^{\circ}$C + 273,15 = 45 $^{\circ}$C + 273.15=318,15 K [/tex]
Now, we clear the Ideal Gas Equation for P, and solve it:
[tex]P*V=n*R*T \\ \\ P= \frac{n*R*T}{V}= \frac{(0,5mol)*(0,082 \frac{L*atm}{mol*K})* (318,15 K)}{18 L}= 0,725 atm [/tex]
Have a nice day!
