Question

A 8.55 L container holds a mixture of two gases at 39 °C. The partial pressures of gas A and gas B, respectively, are 0.348 atm and 0.813 atm. Of 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

`\boxed{\text{1.790 atm}}`

Step-by-step explanation:

To calculate the partial pressure of the third gas, we can use the Ideal Gas Law:

pV = nRT

Data:

V = 8.55 L

n = 0.210 mol

T = 39 °C

Calculations:

T = (39 + 273.15) K = 312.15 K

`\begin{array}{rcl}p * 8.55 & = & 0.210 * 0.082 06 * 312.15\\8.55p & = & 5.379\\p & = & \textbf{0.6291 atm}\\\end{array}`

According to Dalton's Law of Partial Pressures, each gas exerts its own pressure independently of the others.

`\begin{array}{rcl}p_{\text{tot}}& = & p_{\text{A}} + p_{\text{B}} + p_{\text{C}}\\& = & 0.348 + 0.813 + 0.6291\\& = & \textbf{1.790 atm}\\\end{array}\\\text{The total pressure will become } \boxed{\textbf{1.790 atm}}.`