Question

A 7.70-L container holds a mixture of two gases at 35 °C. The partial pressures of gas A and gas B, respectively, are 0.190 atm and 0.891 atm. If 0.230 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

Answer – We are given, volume, V = 7.70 L , T = 35+273 = 308 K

Partial pressure of gas A = 0.190 atm , partial pressure of B = 0.891 atm

Now we need to calculate the moles of A and B

Using he Ideal gas law

For gas A -

n = PV/RT

= 0.190 atm * 7.70 L/ 0.0821 L.atm.mol^-1.K^-1*308 K

   = 0.0578 moles

For gas B -

n = PV/RT

= 0.891 atm * 7.70 L/ 0.0821 L.atm.mol^-1.K^-1*308 K

   = 0.271 moles

So total moles after added third gas = 0.0578 + 0.271 + 0.230

                                                           = 0.559 moles

Now the total pressure –

P = nRT/V

   = 0.559 moles* 0.0821 L.atm.mol^-1.K^-1 * 308 K / 7.70 L

     = 1.84 atm

The total pressure become 1.84 atm