Question

What would be the pressure of 1.0g of O2 contained in a 4.00L container at 293 K?

Answer 1

Answer: The pressure will be equal to 0.19 atm.

Explanation:

The Ideal Gas Equation states the relationship among the pressure, temperature, volume, and number of moles of a gas.

The equation is:

$PV=nRT$

where P = pressure in atm

           V = volume in L

           n = numbers of moles of gas in mol

           R = universal gas constant = 0.08206 $\frac{L-atm}{mol - K}$

           T = temperature in K

Based on the problem,

mass of O2 = 1.0 g

V = 4.00 L

T = 293 K

mol of O2  = ?

P = ?

We need to calculate the moles of O2 before we can use the Ideal Gas Equation. To solve the number of moles, we use the equation:

$no. of moles = \frac{given\ mass\ (in\ grams) }{molar\ mass\ (in\ \frac{grams}{mole}) }$

The molar mass of O2 is 32 g/mol, therefore,

$no. of moles = \frac{1.0g}{32\frac{g}{mol} }$

no. of moles of O2 = 0.03125 mol.

Now we substitute the values into the Ideal Gas equation:

$P(4.00L) = (0.03125 mol)(0.08206\frac{L-atm}{mol-K} )(293K)$

Solving for P, we will get

$P=\frac{(0.03125mol)(0.08206\frac{L-atm}{mol-K})(293K) }{4.00L} \\\\P= 0.1878 atm.$

In correct significant figures, P is equal to 0.19 atm.