Question #5609e

1 Answer
Jan 2, 2015

I'm assuming that 616 mg is actually 616 mmHg, the gas' pressure.

You can approach this problem by using the combined gas law, since the number of moles does not change.

(P_1V_1)/T_1 = (P_2V_2)/T_2

You know that

V_1 = 322 "mL,
T_1 = 29^@C = 273.15 + 29 =302.15 "K"
P_1 = 616 "mmHg" = 616 "mmHg" * ("1 atm")/(760"mmHg") = 0.810 "atm"

At STP, the temperature is equal to T_2 = 273.15 "K" and pressure is equal to P_2 = 1.00 "atm". Plug all in and get

V_2 = P_1/(P_2) * T_2/T_1 * V_1 = (0.810atm)/(1.00atm) * (273.15K)/(302.15K) * 322mL

V_2 = 236 "mL"