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 0.75 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 elevation is 335.2 K or 62.2 °C

Explanation:

Step 1: Data given

The initial volume = 1.2 L

The initial temperature = 25.0 °C = 298 K

The pressure = 1.0 atm

The pressure reduces to 0.75 atm

The volume increases to 1.8 L

Step 2: Calculate the new temperature

(P1*V1)/T1 = (P2*V2)/T2

⇒with P1 = the initial pressure in the balloon = 1.0 atm

⇒with V1 = the initial volume of the balloon = 1.2 L

⇒with T1 = the initial temperature = 298 K

⇒with P2 = the reduced pressure = 0.75 atm

⇒with V2 = the increased volume = 1.8 L

⇒with T2 = the new volume = TO BE DETERMINED

(1.0 atm * 1.2L) / 298K = (0.75 atm * 1.8 L) / T2

0.004027 = 1.35 / T2

T2 = 1.35 / 0.004027

T2 = 335.2 K

335.2 - 273 = 62.2 °C

The temperature of the air at this elevation is 335.2 K = 62.2 °C

Answer 2

Answer:

$T_2=335.42K=62.27^oC$

Explanation:

Hello,

In this case, by using the general gas law, that allows us to understand the pressure-volume-temperature relationship as shown below:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

Thus, solving for the temperature at the end (considering absolute units of Kelvin), we obtain:

$T_2=\frac{P_2V_2T_1}{P_1V_1}=\frac{1.8L*0.75atm*(25+273.15)K}{1.2L*1.0atm} \\\\T_2=335.42K=62.27^oC$

Best regards.