Question

What temperature (in °C) did an ideal gas shift to if it was initially at -10.0 °C at 4.62 atm and 35.0 L and the pressure was changed to 8.71 atm and the volume changed to 15.0 L

Answer 1

-251.9°C

Step-by-step explanation:

Given data:

Initial volume = 35.0 L

Initial pressure = 4.62 atm

Initial temperature = -10.0 °C (-10.0 +273 = 263.0 K)

Final temperature = ?

Final volume = 15.0 L

Final pressure = 8.71 atm

Solution:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Now we will put the values,

4.62 atm × 35.0 L / 263.0 K = 8.71 atm × 15.0 L /T₂

T₂ = 8.71 atm × 15.0 L × 263.0 K / 4.62 atm × 35.0 L

T₂ = 34360.95 atm.L.K /161.7 atm.L

T₂ = 21.26 K

Kelvin to °C:

21.26 - 273.15 K = -251.9°C