Question

The mass of a hot-air balloon and its cargo (not including the air inside) is 200 kg. The airoutside is at a temperature of 10°C and a pressure of 1 atm = 10^5 N/m2. The volume of the balloon is 400 m^3. Which temperature below of the air in the balloon will allow the balloon to just lift off? (Air density at 10°C is 1.25 kg/m^3.)a. 37°Cb. 69°Cc. 99°Cd. 200°C

Answer 1

Use the ideal gas law for the initial and final conditions.

$\begin{gathered} P_iV_i=n_iR_iT_i \\ P_fV_f=n_fR_fT_f \end{gathered}$

Where P_i = 10^5 N/m^2, V_i = 400 m^3, T = 10 C = 283 K. So, we have the expression about the initial conditions

$\frac{10^5N}{m^2}\cdot400m^3=n_iR_i\cdot283K$

For the second expression, the final pressure and the final volume are the same because the hot-air balloon is open to the atmosphere.

$\frac{10^5N}{m^2}\cdot400m^3=n_fR_fT_f$

Then, use the density formula to find the initial mass and the final mass.

$\begin{gathered} m_i=\rho_iV=\frac{1.25\operatorname{kg}}{m^3}\cdot400m^3=500\operatorname{kg} \\ m_f=m_i-200\operatorname{kg}=500\operatorname{kg}-200\operatorname{kg}=300\operatorname{kg} \end{gathered}$

Now, we find the number of moles (n) by dividing the masses.

$\frac{n_i}{n_f}=\frac{500\operatorname{kg}}{300\operatorname{kg}}\approx1.67$

At last, we divide the initial equations (ideal gas law) to find the final temperature.

$\frac{10^5\cdot400}{10^5\cdot400}=\frac{n_iR_i\cdot283K}{n_fR_fT_f}$

The constant R is the same in both cases and use the ratio of moles.

$\begin{gathered} 1=\frac{1.67\cdot283K}{T_f} \\ T_f=472.61K\approx473 \end{gathered}$

But, in Celsius, the temperature is 200 Celcius.

Therefore, the answer is D.