A sample of 62.3 CM^3 of argon gas at 18°c was contained at a pressure of 155 kPa in a j-shaped tube with Mercury. Later the temperature changed. When the Mercury level was adjusted to give the same pressure of argon, the gas volume changed to 61.2 cm^3. What was the final temperature of the argon?

Answer 1

V1/T1 = V2/T2
so plug in your numbers and find T2

Answer 2

V1/T1=V2/T2. NOTE 62.3cm^3 = 62.3/1000 =0.0623L

Answer 3

V1/T1=V2/T2. NOTE 62.3cm^3 = 62.3/1000 =0.0623L