If a sample of N gas was put into a flexible container at an pressure of 25.5kPa, temperature of 35.9 C and a volume of 500. mL, what would the final temperature of the gas be if the pressure was changed to .625 atm and the volume was change to 625mL?

1 Answer

Answer:

#705 ^@"C"#

Explanation:

First look at the things that need to be converted

#35.9 ^@"C" = "(35.9 + 273.15) K" = "309.0 K"#

#"25.5 kPa = 0.2467 atm"#

#"500 mL = 0.500 L"#

#"625 mL = 0.625 L"#

#(P_1V_1)/T_1 = (P_2V_2)/T_2#

Plug in the variables

#("0.2467 atm" xx "0.500 L")/("309.0 K") = ("0.625 atm" xx "0.625 L")/T_2#

#T_2 = 978.5K#

or #T_2 = (978.5 - 273.15) ^@"C" = 705 ^@"C"#