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
Another turor helped me solve this problem I am ju sure how they converted presuure to...
A helium balloon has a volume of 3.22 L at 21.4C and a pressure of 755mm Hg. a.) 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? b.) If the same balloon is placed in a vacuum chamber and the pressure is reduced to .551 atm, what will the new volume of the balloon in liters, assuming the temperature remains constant?
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