Question

The pressure in a container has decreased by 10 times, while the volume increased by 5 times from the original volume. The temperature now reads -123.15 °C. What was the starting temperature when I left (in °C)?

Answer 1

Answer: The starting temperature when I left in °C is 26.70

Step-by-step explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of gas = p

`P_2` = final pressure of gas =
`(p)/(10)`

`V_1` = initial volume of gas = v

`V_2` = final volume of gas =
`5* v`

`T_1` = initial temperature of gas = ?

`T_2` = final temperature of gas =
`-123.15^oC=273+(-123.15)=149.85K`

Now put all the given values in the above equation, we get:

`(p* v)/(T_1)=((p)/(10)* 5* v)/(149.85K)`

`T_1=299.70K=(273-299.70)^0C=-26.70^0C`