High-Altitude Research. A large research balloon containing 2.00 x 10³ m³ of helium gas at 1.00 atm and a temperature of 15.0°C rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm (Fig. P19.50). Assume the helium behaves like an ideal gas and the balloon’s ascent is too rapid to permit much heat exchange with the surrounding air. (a) Calculate the volume of the gas at the higher altitude. (b) Calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher attitude ?

The volume is given as,$v_1=2.00\times {10}^3{m}^3$

Pressure as$P_1=1.00atm$

Temperature asT = 15 ℃

Take pressure at higher altitude as$P_2=0.900atm$

In adiabatic process,

$P_1{v_1}^{\gamma }=P_2{v_2}^{\gamma }$

In ideal monoatomic gas,$\gamma =\frac{5}{3}=1.67$

Take$V_2={\left(\frac{P_1}{P_2}\right)}^{\frac{1}{\gamma }}{V}_1$

Put all the values,

$V_2={\left(\frac{1.00}{0.90}\right)}^{\frac{3}{5}}\times 2.00\times {10}^3=2.131\times {10}^3{m}^3$

Thus, the volume of the gas at the higher altitude is$2.131\times {10}^3{m}^3$ .

$$\begin{gathered} T_1{V_1}^{\gamma -1}=T_2{V_2}^{\gamma -1} \\ {T}_2={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}{T}_1 \end{gathered}$$

Put all the values,

$$\begin{gathered} T_2={\left(\frac{2.00\times {10}^3}{2.131\times {10}^3}\right)}^{\frac{5}{3}}\times 15.0 \\ {=}_1{\left(\frac{2.00\times {10}^3}{2.131\times {10}^3}\right)}^{\frac{2}{3}}\times 15.0 \\ =14.38 \end{gathered}$$

Thus, the temperature of the gas at the higher altitude is14.38℃

For adiabatic process $Q=0$and$∆U=-W$

The work done is,

$$\begin{gathered} W=\frac{1-\gamma }{\gamma -1}\left(P_1{V}_1-P_2{V}_2\right) \\ =\frac{1}{{\displaystyle \frac{5}{3}}-1}\times \left[\left(1.00\times 2.00\times {10}^3\right)-\left(0.90\times 2.131\times {10}^3\right)\right] \\ =\frac{3}{2}\times \left[\left(2.00\times {10}^3-1.92\times {10}^3\right)\right] \\ =0.12\times {10}^{3}J \end{gathered}$$

Thus, the change in internal energy of the helium as the balloon rises to the higher altitude is $-0.12\times {10}^{3}J$.