chemistry question

Question

A gas-filled weather balloon has a volume of 51.0

L at ground level, where the pressure is 765 mmHg and the temperature is 21.6 °C. After being released, the balloon rises to an altitude where the temperature is −5.43 °C and the pressure is 0.0695 atm. What is the weather balloon's volume at the higher altitude?

Answer 1

1st!! you will need to convert units of temperature (C to K) and pressure (mmHg to Atm) why? So that the dimensional analysis is correct (keeping all your units in the same system) and also the relationship between temperature and kinetic energy is only kept in degrees K.

V₁= 51L, T₁=21.6C(294.6 K), P₁=765mmHg(1.007 Atm)

V₂=? , T₂=-5.43C(267.57K), P₂=0.0695 Atm

Using the combined gas law:

P₁*V₁ P₂V₂

--------- = --------

T₁ T₂

Solving for V₂ = (P₁ * V₁ * T₂) / (T₁ * P₂ )

V₂ = (1.007 Atm * 51 L * 267.57 K)

----------------------------------------- = through dimensional analysis the only remaining unit is L for volume

(294.6 K * 0.0695 Atm)

V₂= 671.15 L, this answer makes sense, since pressure decreased we should end with a larger volume than we started with.

Answer 2

Combined Gas Law

P1V1/T1 = P2V2/T2

Use this equation to solve for V2 BUT be sure to get all units to be consistent.

P1 = 765 mm Hg

V1 = 51.0 Liters

T1 = 21.6C + 273 = 294.6K

P2 = 0.0695 atm x 760 mm Hg/atm = 52.82 mm Hg

V 2 = ?

T2 = -5.43C + 273 = 267.6K

plug and chug