Question

Another turor helped me solve this problem I am ju sure how they converted presuure to atm?

5. (4 points) A helium balloon has a volume of 3.22 L at 21.4C and a pressure of 755 mm Hg. Vi - 3.220 d. If the balloon is immersed in a cold temperature bath of -63.7C and the pressure remains constant, what will be the new volume in liters of the balloon? P = 755 mmHg = 019- Ti = 21 4C-294.4K P2F2= Pike 1 = 1 TZT -630=209.3k T2 Ti 3.23 = 3.227 209, 3 = 2.296 29119 294.4 b. If the same balloon is placed in a vacuum chamber and the pressure is reduced to 0.551 atm, what will be the new volume of the balloon in liters, assuming the temperature remains constant? 5.80 L

Answer 1

a. From Charles' law:

Volume / temperature = constant, at constant pressure and temperature is in K.

T₁ = 273 + 21.4 = 294.4 K

T₂ = 273 - 63.7 = 209.3 K

Assuming the final Volume is V:

3.22 / 294.4 = V / 209.3

Solving, V = 2.29 L

b. Pressure in atm = pressure in torr / 760 = 755 / 760 = 0.993 atm

From Boyle's law at constant pressure:

Pressure * Volume = constant

0.993 * 3.22 = 0.551 * V

Solving, V = 5.80 L