Question

A toy helium balloon is initially at a temperature of T = 24o C. Its initial volume is 0.0042 m3 (It is a 10 cm radius sphere). Assume that the pressure in the balloon always equals atmospheric pressure, 101.3 kPa. 1)The balloon is lying in the sun, which causes the volume to expand by 13%. What is the new temperature, T? T =

Answer 1

Answer:

$T_{f} = 335.780\,K\,(62.630\,^{\textdegree}C)$

Explanation:

Let assume that air behaves ideally. The equation of state of ideal gases is:

$P\cdot V = n\cdot R_{u}\cdot T$

Where:

$P$ - Pressure, in kPa.

$V$ - Volume, in m³.

$n$ - Quantity of moles, in kmol.

$R_{u}$ - Ideal gas constant, in $\frac{kPa\cdot m^{3}}{kmol\cdot K}$.

$T$ - Temperature, in K.

Since there is no changes in pressure or the quantity of moles, the following relationship between initial and final volumes and temperatures is built:

$\frac{V_{o}}{T_{o}} = \frac{V_{f}}{T_{f}}$

The final temperature is:

$T_{f} = \frac{V_{f}}{V_{o}}\cdot T_{o}$

$T_{f} = 1.13\cdot (297.15\,K)$

$T_{f} = 335.780\,K\,(62.630\,^{\textdegree}C)$

Answer 2

Answer:

The new temperature of the balloon is 27.12⁰C

Explanation:

Given;

initial temperature of helium gas in balloon, T₁ = 24⁰ C

initial volume of the gas, V₁ =  0.0042 m³

pressure in the balloon, P = 101.3 kPa

The sun caused the balloon to expand by 13 % of the original volume; this implies that the original volume increased by 13%.

The new temperature, T₂ is calculated using general gas law;

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Since the pressure in the balloon is always equal, then P₁ = P₂

$\frac{V_1}{T_1} = \frac{V_2}{T_2}\\\\T_2 = \frac{T_1V_2}{V_1}\\\\T_2 =\frac{24(0.0042\ +\ 0.13*0.0042)}{0.0042} \\\\T_2 = \frac{24(0.004746)}{0.0042} \\\\T_2 = 27.12 \ ^oC$

Therefore, the new temperature of the balloon is 27.12⁰C