4.A stoppered test tube traps 26 cm3 of air at a pressure of 1 atm and a tempera

Question

4.A stoppered test tube traps 26 cm3 of air at a pressure of 1 atm and a temperature of 17 °C. The stopper has a diameter of 1.4 cm and will pop off if the net upward force is 14 N. Assume the air surrounding the test tube is always at 1 atm.
A.What is the mass of the air in the test tube?
(include units with answer)
B.To what temperature (absolute) would you have to heat the trapped air to pop off the test tube stopper?
(include units with answer)
C.If the temperature remained a constant 17 °C, what is the total mass of air the test tube can hold before the top will pop off?
(include units with answer)

Explanation / Answer

1 atm = 1.013 e5 Pa

Area A = pi r^2 = 3.14 * 0.007*0.007

A = 1.54 *10^-4 m^2

Force in Downward direction = PA

Fd = 1.013 *e 5 * 1.54e -4

Fd = 15.59 N

net Force Fup = 14+15.59 = 29.59 N

mass m = 29.59/9.8 1

mass = 3.017 Kgs

--------------------------------

pressure Inside = F/A

P = 29.59/1.54 e-4

P = 1.92 e 5 Pa    or 1.92 atms
-----------------------------

Initial Temp Ti = 17+273 = 290 K

use P1/P2 = T1/T2

T2 = 290 * 1.92/(1.013)

T2 = 550 K or 277deg C

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