Question

Ly Give Up? ATVV T Resources Hint Check Answer Question 16 of 34> Attempt 2 A 7.90 L container holds a mixture of two gases at 19 °C. The partial pressures of gas A and gas B, respectively. are 0.200 atm and 0.692 atm. If 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become? Ptal atm

Answer 1

Answer-

Given,

Volume = 7.90 L

Temperature = 19^$\degree$C or 292.15 K [0°C + 273.15 = 273.15K]

Partial Pressure of A = 0.200 atm

Partial Pressure of B = 0.692 atm

Moles of Third Gass C = 0.210 mol

Total Pressure = ?

We know that,

P_total = P_A + P_B + P_C

where, P_A , P_B , P_C are partial pressures

Also,

PV = nRT

where, P= Pressure in atm

V = Volume in L

n = Moles

R = Gas Constant (0.082057 L.atm.mol^-1.K^-1)

T = Temperature in K

So, partial Pressure of Third Gas ( P_C ),

P_C = ( 0.210 mol * 0.082057 L.atm.mol^-1.K^-1 * 292.15 K ) / 7.90 L

P_C = 0.637 atm

So, P_total = P_A + P_B + P_C

P_total = 0.200 atm + 0.692 atm + 0.637 atm

P_total = 1.53 atm [Answer]