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

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

Answer 2

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,

$\frac{V_1}{V_2}=\frac{T_1}{T_2}$

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.

$\frac{0.456L}{3.4L}=\frac{T_1}{338K}$

$T_1=45.33K$

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