Question

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an elevation where the pressure is 0.73 atm, then the new volume is 1.8 L. What is the temperature (in °C) of the air at this elevation?

Answer 1

Answer:

The temperature of the air at this given elevation will be 53.32425°C

Explanation:

We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.

Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm  ; Final Volume = 1.8 L ; Final pressure = 0.73 atm  

We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T$_2$ ),

P$_1$V$_1$ / T$_1$ = P$_2$V$_2$ / T$_2$,

T$_2$ = P$_2$V$_2$T$_1$ / P$_1$V$_1$,

T$_2$ = 0.73 atm $*$ 1.8 L $*$ 298.15 K / 1 atm $*$ 1.2 L = ( 0.73 $*$ 1.8 $*$ 298.15 / 1 $*$ 1.2 ) K = 326.47425 K,

T$_2$ = 326.47425 K = 53.32425 C