A 1.37 L vessel contains He at a temperature of 24.5 °C and a pressure of 205 mmHg. A 721 mL vessel contains Ne at a temperature of 36.2 °C and a pressure of 0.185 atm. Both of these gases are placed in a 2.00 L vessel at 302 K. What is the final pressure (in atm) in the 2.00 L vessel?

.253 atm

Explanation:

You would use Dalton's law of partial pressures which is the sum of all gas pressures will be total pressure. But first you must use combined gas law to determine the "new" pressures in the 2L vessel at the 302 K. the formula is  P1 x V1 / T1 = P2 x V2 / T2, where temperature MUST be in Kelvin. Convert your 2 temperatures to Kelvin by adding 273.15). You will then do the combined gas law for the Helium using the values given calculating as P2 your unknown, 2.00L & 302K are  V2 & T2 respectively (and for the Ne equation that you will do after you have calculated this one). Make sure to convert 205 mmHg to atm (.2697atm), and solve for the unknown pressure:      .2697atm x 1.37L x 302K / 297.65K x 2L = .1874atm (when you cancel the units you are left with atm, which is what the final answer is asking for). Now combined gas law for the Neon using same formula and your V2 & T2 from previous problem (the 2 gasses are going to the SAME new temp and volume). Convert milliliters to liters by simply moving the decimal to the left 3 times to get 0.721L. so 0.185atm x 0.721L x 302K / 309.35K x 2.00L = 0.06510atm. NOW use Dalton's law of partial pressures, just add them up: .1874atm + 0.06510atm = 0.2525atm BUT dont forget significant figures, in this case you need 3 so take the last number and apply rounding rules (meaning 252 are significant but there is a 5 after the last significant figure so the second 2 gets rounded to 3)

2b2t

explanation:

2b2t