Question

A 8.05 L container holds a mixture of two gases at 11 °C. The partial pressures of gas A and gas B, respectively, are 0.417 atm and 0.773 atm. If 0.110 mol of a third gas is added with no change in volume or temperature, what will the total pressure become? Press

Answer 1

partial pressure of A P_A =0.417 atm

partial pressure of B P_B =0.773 atm

volume V =8.05 L

temperature T=11⁰C =273+11 =284 K

moles of gas C =0,110moles

since there is no change in volume or temperature therefore we can assume the gas C to be ideal in nature and ideal gas equation can be used to calculate the pressure of Gas C

PV=nRT

P_c = nRT/V

P_c = 0.110 moles *0.0826 atm L K^-1 mol-1 *284 k/ 8.05l

P_{c =} 2.580 atm /8.05

P_c =0.32 atm

according to Dalton's Law of Partial Pressures.: in a gaseous mixture, the partial pressure of a gaseous component is the same as the pressure it would exert if it alone occupied the container.

P_total =P_A+P_B+P_C

P_total = 0.417 atm +0.773 atm +0.32 atm

P_total =1.51 atm