Answer:
[tex]T_2=335.42K=62.27^oC[/tex]
Explanation:
Hello,
In this case, by using the general gas law, that allows us to understand the pressure-volume-temperature relationship as shown below:
[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]
Thus, solving for the temperature at the end (considering absolute units of Kelvin), we obtain:
[tex]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[/tex]
Best regards.
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
