Question

At what temp will a gas be at if you allow it to expand from an original 456 mL to 65°C to 3.4 L

Answer 1

Answer : The initial temperature of gas will be, 45.33 K

Explanation :

Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

`V\propto T`

or,

`(V_1)/(V_2)=(T_1)/(T_2)`

where,

`V_1` = initial volume of gas = 456 ml = 0.456 L

conversion used : (1 L = 1000 ml)

`V_2` = final volume of gas = 3.4 L

`T_1` = initial temperature of gas = ?

`T_2` = final temperature of gas =
`65^oC=273+65=338K`

Now put all the given values in the above formula, we get the initial temperature of the gas.

`(0.456L)/(3.4L)=(T_1)/(338K)`

`T_1=45.33K`

Therefore, the initial temperature of gas will be, 45.33 K

Answer 2

We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
V₁T₂ = V₂T₁
Substituting the known values,
(0.456 L)(65 + 273.15) = (3.4 L)(T₁)
T₁ = 45.33 K