Question

Question #5609e

Answer 1

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"`