Question

A flexible .556 L container holds air at a temperature of 27.3 degrees C. If the container is cooled so that the new volume of the gas is .214 L, what is the new temperature of the air in Celcius?

Answer 1

`115.64K` or in Celcius, `-157.51^@C`.

Explanation:

Charle's Law `(V1)/(T1)=(V2)/(T2)`
`V`'s unit must be in liters (L) and Kelvin (K)=`C^@+273.15` for `T`
`V_1=0.556L`
`T_1=27.3^@C+273.15=300.45K`
`V_2=0.214L`
`T_2=?`

Now, to solve for `T_2` multiply by the appropriate variables so that T2 ends up on the top and by itself.
`(cancel(T_1)*T_2)/cancel(V_1)*cancel(V_1)/cancel(T_1)=(V_2)/cancel(T_2)*(T_1*cancel(T_2))/(V_1)` simply `->` `T_2=(V_2)/(V_1)*T_1`

`T_2=(0.214cancel(L))/(0.556cancel(L))*300.45K`
`T_2=?`