Question

A 1.20 L weather balloon on the ground has a temperature of 25.0°C and is at atmospheric pressure (1.00 atm). When it rises to an elevation where the pressure is 0.720 atm, then the new volume is 1.80 L. What is the temperature (in °C) of the air at this elevation?

Answer 1

$\begin{aligned} V_1 &= 1.2L\\ T_1 &= 25°C = 273+ 25 = 298K\\ P_1 &= 1atm\\ P_2 &= 0.72atm\\ V_2 &= 1.8L\\ T_2 &= x°C = (273+x)\\ \textsf{From t}&\textsf{he general gas equation,}\\ \dfrac{P_1V_1}{T_1} &= \dfrac{P_2V_2}{T_2}\\ \\ \dfrac{1×1.2}{298} &= \dfrac{1.8× 0.72}{x+273}\\ \\ \dfrac{1.2}{298} &= \dfrac{1.296}{x+273}\\ \\ 1.2(x+273) &= 298× 1.296\\ 1.2x + 327.6 &= 386.208\\ 1.2x &= 58.608\\ x &= 48.84°C \end{aligned}$