To what temperature would you have to heat the trapped air in order to "pop off" the stopper? assume the air surrounding the test tube is always at a pressure of 1.00 atm.

Area of stopper = π * 0.011^2 Force pushing DOWN on top = 1.013 * 10^5 * π * 0.011^2 = 38.507 N 15.0 = Force up – 38.507 Force up = 15.0 + 38.507 = 53.507 N Pressure on inside surface of stopper = Force up ÷ area Pressure = 53.507 ÷ (π * 0.011^2) = 1.4076 * 10^5 Initial Temperature = 18 + 273 = 291°K 1.013 * 10^5 / 1.4076 * 10^5 = 291 / T T = 404°K – 273 = 131°C

Where's the pictures ?

Iam not sure butch i thinnk it is ab