Question

A weather balloon at Earth’s surface has a

Answer 1

The temperature : 263.016 K

Further explanation

Combined with Boyle's law and Gay Lussac's law  

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

P1 = initial gas pressure (N/m² or Pa)  

V1 = initial gas volume (m³)  

P2 = gas end pressure  

V2 = the final volume of gas  

T1 = initial gas temperature (K)  

T2 = gas end temperature  

P1=760 mmHg

V1= 4 L

T1 = 275 K

P2=704 mmHg

V1=4.13 L

$\tt \dfrac{760\times 4}{275}=\dfrac{704\times 4.13}{T_2}\\\\T_2=263.016~K$