91Ó°ÊÓ

1. The combined weight of the envelope and the basket of air-balloon,$W_{combined}=2.45\text{ k±·}$
2. Capacity of the envelope,$V_{envelop}=2.18×{10}^3{\text{″¾}}^3$
3. Lifting capacity force applied on the envelope to get fully inflated envelope (in addition to balloon’s weight),$F_{net}=2.67\text{kN}$
4. Temperature of the surrounding air,$T_{surrounding}=20.0\text{ }°\text{C}$
5. Weight per unit volume,$F_C/V=11.9{\text{N/m}}^3$
6. Molecular mass of the weight,$M=0.028\text{ k²µ}/\text{mol}$
7. Pressure on the weight,$P=1\text{ a³Ù³¾}$

The buoyant force on a body is given by the pressure applied on the body by a fluid. Because the pressure increases as the depth increase, the pressure on the bottom of an object is always larger than the force on the top, hence,a net upward force acts on the body. As we are given a lifting force, thus the net force is in an upward direction which implies the force on the envelope is greater than the combined weight.

The ideal gas equation,

$PV=nRT$ (i)

where, P is the pressure applied on the gas, V is the volume of the gas, n is the number of moles of the gas, R is the gas constant, T is the temperature of the gas.

The buoyant force on a body due to an applied pressure,

$F=\mathrm{ÒÏ}\text{Vg}$(ii)

where, P is the density of the body, v is the volume of the body, g is the acceleration due to gravity.

The force applied on the envelope can be given using equation (ii) as follows:

$F_b={\mathrm{ÒÏ}}_{c}Vg$

where,${\mathrm{ÒÏ}}_c$is the density of the envelope.

Similarly, the total force is given as:

$F_g=W+{\mathrm{ÒÏ}}_{h}gV$

where, W is the combined weight of the body,${\mathrm{ÒÏ}}_h$is the density of the hot-air balloon due to the applied pressure on the body.

Thus, the net lifting force value acting on the balloon is the difference between the two above forces that is given as follows:

$\begin{array}{l}{F}_{net}=F_b−F_g\\ F_{net}={\mathrm{ÒÏ}}_{c}Vg−(W+{\mathrm{ÒÏ}}_{h}gV)\\ {\mathrm{ÒÏ}}_{h}gV={\mathrm{ÒÏ}}_{c}Vg−W−F_{net}\\ {\mathrm{ÒÏ}}_{h}g=\frac{({\mathrm{ÒÏ}}_{c}Vg−W−F_{net})}{V}\end{array}$

For the given values-

$\begin{array}{c}{\mathrm{ÒÏ}}_{h}g=\frac{((11.9{\text{F/m}}^3×2.18×{10}^3{\text{m}}^3)−2.45×{10}^3\text{N}−2.67×{10}^3\text{N})}{2.18×{10}^3{\text{m}}^3}\\ =9.55\text{ N}/{\text{m}}^3\end{array}$

Now, from the ideal gas equation, we get the pressure volume ratio as follows:

$\frac{P}{RT}=\frac{n}{V}$

Multiplying both the sides of the above equation by the molar weight, we get the temperature of the gas using the given data as follows

$\begin{array}{l}\frac{PMg}{RT}=\frac{nMg}{V}\left(âˆ\mu \text{Densityofabody,}\mathrm{ÒÏ}=\frac{\text{M}}{\text{V}}\right)\\ \frac{PMg}{RT}=n{\mathrm{ÒÏ}}_{h}g\\ T=\frac{PMg}{Rn\mathrm{ÒÏ}g}\end{array}$

For the given values-

$\begin{array}{c}T=\frac{1.01×{10}^5\text{Pa}×0.028\text{kg}×9.8{\text{m/s}}^2}{8.31{\text{m}}^3\text{ P²¹/°­.³¾´Ç±ô}×1\text{mol}×9.55{\text{N/m}}^3}\\ =349.2\text{‿é}\end{array}$

The temperature for enclosed air will be $349.2\text{‿é}$.