Question

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

Answer 1

7,859.9829 kPa

Step-by-step explanation:

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:

`P=(n*R*T)/(V)` 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?

`moles=(94.6 g*1 mole)/(44 g)`

moles= 2.15

Then you know the following data:

• n= 2.15 moles
• R= 0.082
  `(atm*L)/(mol*K)`
• T= 57 C= 330 K (0 C= 273 K)
• V= 0.75 L

Replacing in Equation (A):

`P=(2.15 moles*0.082 (atm*L)/(mol*K) *330 K)/(0.75L)`

Solving:

P= 77.572 atm

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