Question

A helium balloon at room temperature ( 25 degree ) occupies a volume of 2.0 L. When the balloon is expanded to 5.0 L, the balloon will finally pop . If the pressure is not changed, at what temperature will this occure

Answer 1

745.4K ~ 472.3 C

Step-by-step explanation:

This is an Ideal Gas Law problem where we have to manipulate the equation a bit. Let's start with the basic:

PV = nRT will be used for both the initial and final, so we will rearrange this problem to state:

(V(initial))/(T(Initial)) = nR/P

Since we know that the pressure, number of moles of He, and ideal gas constant (R) remain the same from start to finish so we can write the problem as such:

(V(initial))/(T(Initial)) = nR/P = (V(final))/(T(final))

or

(V(initial))/(T(Initial)) = (V(final))/(T(final))

Now lets define some of these values:

T(initial) = 25degree (assuming degrees Celsius) ~ 298.15K

V(initial) = 2.0L

V(final) = 5.0L

T(final) = ?

Since we are solving for T(final) let's rearrange the problem once more to be solving for T(final):

T(final) = (V(final)T(Initial))/V(initial)

Now plug in your values:

T(final) = (5.0L*298.15K)/(2.0L) ~ 745.4K ~ 472.3degrees Celsius