Question

A 9.80-L container holds a mixture of two gases at 53

Answer 1

given

volume of container = 9.8

temperature = 53 C = 53 + 273 = 326 K

now

total pressure = partial pressure of A (PA) + partial pressure of B (PB)

so

total pressure = 0.165 + 0.829

total pressure (P) = 0.994 atm

we know that

PV = nRT

so

0.994 x 9.80 = n x 0.0821 x 326

n = 0.364

so

the total moles = 0.364

now

0.18 moles of third gas is added

so

new total moles = 0.364 + 0.18 = 0.544

also given

volume and temperature remains constant

so use

PV = nRT

P x 9.8 = 0.544 x 0.0821 x 326

P = 1.4856

so the total pressure becomes 1.4856 atm