Question

a 9.45 l container holds a mixture of two gases at 37°c. the partial pressures of gas a and gas b, respectively, are 0.235 atm and 0.612 atm. if 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
p_{total} = atm

Explanation:

Step1: Convert temperature to Kelvin

$$T=(37 + 273.15)\text{ K}=310.15\text{ K}$$

Step2: Calculate the partial pressure of the third gas

Use the ideal gas law \(PV = nRT\). Rearranged for \(P\), we have \(P=\frac{nRT}{V}\).
Given \(n = 0.210\text{ mol}\), \(R=0.0821\text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\), \(V = 9.45\text{ L}\), and \(T = 310.15\text{ K}\)
$$P_{C}=\frac{0.210\times0.0821\times310.15}{9.45}\text{ atm}$$
$$P_{C}=\frac{5.34}{9.45}\text{ atm}\approx0.565\text{ atm}$$

Step3: Calculate the total pressure

By Dalton's law of partial pressures \(P_{total}=P_{A}+P_{B}+P_{C}\)
Given \(P_{A}=0.235\text{ atm}\), \(P_{B}=0.612\text{ atm}\), and \(P_{C} = 0.565\text{ atm}\)
$$P_{total}=(0.235 + 0.612+0.565)\text{ atm}$$
$$P_{total}=1.412\text{ atm}$$

Answer:

\(1.41\) atm