Question

A 7.70 L 7.70 L container holds a mixture of two gases at 47 ° C. 47 °C. The partial pressures of gas A and gas B, respectively, are 0.344 atm 0.344 atm and 0.893 atm. 0.893 atm. If 0.190 mol 0.190 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

Answer:

1.884 atm

Explanation:

We have at first, a mix of two gases A and B, with a pressure and temperature. The total pressure of this mix is the sum of both pressures fo these gases so:

P = 0.344 + 0.893 = 1.237 atm

Now after this, a third gas is added but there is no change in temperature or volume, which means that this gas may have a behavior of ideal gas, therefore, we can use the expression for ideal gas which is:

P = nRT/V

R is the gas constant, in this case is 0.082 L atm/ K mol

Replacing all the values that we have of volume and temperature, we can calculate the pressure that this gas exert:

P = 0.19 * 0.082 * (47+273.15) / 7.7

P = 0.647 atm

Now the total pressure, we add this value and we have:

P' = 1.237 + 0.647

P' = 1.884 atm