Question

A weather balloon at Earth’s surface has a

volume of 4 L at 275 K and 760 mm Hg. If
the balloon is released and the volume reaches
4.13 L at 704 mm Hg, what is the temperature?
Answer in units of K.

Answer 1

The temperature : 263.016 K

Further explanation

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

`\tt (P_1.V_1)/(T_1)=(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 (760* 4)/(275)=(704* 4.13)/(T_2)\\\\T_2=263.016~K`