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  

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

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

V1 = initial gas volume (m³)  

P2 = 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

$\tt \dfrac{251\times 1.9}{255}=\dfrac{205\times 1.65}{T_2}\\\\T_2=\dfrac{205\times 1.65\times 255}{251\times 1.9}=180.863~K$