Question

A 8.06 g piece of solid CO 2 is allowed to sublime in a balloon. The final volume of the balloon is 1.00 L at 300.0 K. What is the pressure of the gas?

Answer 1

Answer:

The pressure of the gas is 4.428 atm.

Explanation:

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

Considering a certain amount of ideal gas confined in a container where the pressure, volume and temperature can vary, but keeping the mass constant, that is, without altering the number of moles, the pressure, P, the temperature can be related, T and the volume, V, of an ideal gas using the ideal gas law:

P*V = n*R*T

where R is the ideal gas constant, and n is the number of moles of the gas.

In this case,  to know n you must know the molar mass of the CO₂ compound. Being:

• C: 12 g/mole
• O: 16 g/mole

then, the molar mass of CO₂ is: 12 g/mole + 2*16 g/mole= 44 g/mole

Then you apply a rule of three: if 44 grams of CO₂ are present in 1 mole, 8.06 grams in how many moles are they?

$moles of CO_{2} =\frac{8.06 grams*1 mole}{44 grams}$

moles of CO₂= 0.18 moles

Then, you know:

• P: ?
• V: 1 L
• n: 0.18 moles

• R= 0.082$\frac{atm*L}{mol*K}$

• T= 300 K

Replacing in the equation of the ideal gas law:

P* 1 L= 0.18 moles* 0.082$\frac{atm*L}{mol*K}$ * 300 K

Solving:

$P=\frac{0.18 moles* 0.082 \frac{atm*L}{mol*K} *300 K}{1 L}$

P= 4.428 atm

The pressure of the gas is 4.428 atm.