Question

A sample of xenon gas occupies 1.90L at 251 kPa and -18°C. What was the temperature of the xenon if the gas originally measured 1.65 L at 205 kPa?

Answer 1

The temperature of the xenon : 180.863 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 = final gas pressure

V2 = finalgas volume

T1 = initial gas temperature (K)

T2 = final gas temperature

P1=251 kPa

T1=-18+273=255 K

V1=1.9 L

P2=205 kPa

V2=1.65 L

`\tt (251* 1.9)/(255)=(205* 1.65)/(T_2)\\\\T_2=(205* 1.65* 255)/(251* 1.9)=180.863~K`