Answer :
Answer:
[tex]T_{f} = 335.780\,K\,(62.630\,^{\textdegree}C)[/tex]
Explanation:
Let assume that air behaves ideally. The equation of state of ideal gases is:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex]
Where:
[tex]P[/tex] - Pressure, in kPa.
[tex]V[/tex] - Volume, in m³.
[tex]n[/tex] - Quantity of moles, in kmol.
[tex]R_{u}[/tex] - Ideal gas constant, in [tex]\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex].
[tex]T[/tex] - 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:
[tex]\frac{V_{o}}{T_{o}} = \frac{V_{f}}{T_{f}}[/tex]
The final temperature is:
[tex]T_{f} = \frac{V_{f}}{V_{o}}\cdot T_{o}[/tex]
[tex]T_{f} = 1.13\cdot (297.15\,K)[/tex]
[tex]T_{f} = 335.780\,K\,(62.630\,^{\textdegree}C)[/tex]
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;
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
Since the pressure in the balloon is always equal, then P₁ = P₂
[tex]\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[/tex]
Therefore, the new temperature of the balloon is 27.12⁰C