Answer to Question #108853 in General Chemistry for ali

Question

A 50.0 g sample of solid CO2 (dry ice) is added at -100 °C to an evacuated (all of the gas removed) container with a volume of 5.0 L. If the container is sealed and then allowed to warm to room temperature 125 °C so that the entire solid CO2 is converted to a gas, what is the pressure inside the container?

Answer 1

Firstly, I don't understand what the last temperature is, because 125 °C isn't room temperature. That's why I calculated the pressure for two cases:

1)25°C

n(CO2) = m(CO2)/Mr(CO2) = 50/44 = 1.136 mol

Volume of 1 mol at standard temperature and pressure(STP - 273 ^oK, 1 atm) is 22.4L; volume of 1.136 mol at STP:

V=1.136*22.4=25.45 L

General gas law states:

p1*V1/T1=p2*V2/T2;

p1= 1atm; V1=25.45L ; T1=273 K;

V2= 5L; T2=273+25=298 K; p2- ?

p2=p1*V1*T2/(V2*T1) = 1*25.45*298/(5*273) = 5.556 atm;

2) 125 °C

n(CO2) = m(CO2)/Mr(CO2) = 50/44 = 1.136 mol

Volume of 1 mol at standard temperature and pressure(STP - 273 oK, 1 atm) is 22.4L; volume of 1.136 mol at STP:

V=1.136*22.4=25.45 L

General gas law states:

p1*V1/T1=p2*V2/T2;

p1= 1atm; V1=25.45L ; T1=273 K;

V2= 5L; T2=273+125=398 K; p2- ?

p2=p1*V1*T2/(V2*T1) = 1*25.45*398/(5*273) = 7.42 atm;

Answer: 1) if the temperature is 25°C, the pressure is 5.556 atm;

2) if the temperature is 125°C, the pressure is 7.42 atm;