a balloon is filled with 80.5L of helium at sea level. if the balloon rises to a height of the atmosphere where the pressure is .55atm what will the new volume of the balloon be?
+33665 Assuming the temperature is the same (which it won't be in reality!) then pressure*volume is constant, so:
80.5*1 = V*0.55 (Taking pressure at sea level as 1 atm)
V = 80.5/0.55 L
$${\mathtt{V}} = {\frac{{\mathtt{80.5}}}{{\mathtt{0.55}}}} \Rightarrow {\mathtt{V}} = {\mathtt{146.363\: \!636\: \!363\: \!636\: \!363\: \!6}}$$ L
V ≈ 146.4L
+33665 Assuming the temperature is the same (which it won't be in reality!) then pressure*volume is constant, so:
80.5*1 = V*0.55 (Taking pressure at sea level as 1 atm)
V = 80.5/0.55 L
$${\mathtt{V}} = {\frac{{\mathtt{80.5}}}{{\mathtt{0.55}}}} \Rightarrow {\mathtt{V}} = {\mathtt{146.363\: \!636\: \!363\: \!636\: \!363\: \!6}}$$ L
V ≈ 146.4L
