What pressure in kPa is exerted by 94.60 g CO2 at 57.00 ℃ in 0.7500 L?

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables where R is the molar constant of the gases:

P * V = n * R * T

Then the pressure can be calculated as:

[tex]P=\frac{n*R*T}{V}[/tex] Equation (A)

In this case R is a known constant and the volume and temperature are given as data. The number of moles n can be calculated from the molar mass. You know:

C: 12 g/mole
O: 16 g/mole

Then the molar mass of CO₂ is:

CO₂= 12 g/mole + 2*16 g/mole= 44 g/mole

Now you can apply the following rule of three: if 44 g are contained in 1 mole of CO₂, 94.6 g in how many moles are they?

[tex]moles=\frac{94.6 g*1 mole}{44 g}[/tex]

moles= 2.15

Then you know the following data:

n= 2.15 moles
R= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]
T= 57 C= 330 K (0 C= 273 K)
V= 0.75 L

Replacing in Equation (A):

[tex]P=\frac{2.15 moles*0.082 \frac{atm*L}{mol*K} *330 K}{0.75L}[/tex]

Solving:

P= 77.572 atm

Knowing that 1 atm is equal to 101.325 kPa, then 77.572 atm is 7,859.9829 kPa