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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

Respuesta :

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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 : 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.

[tex]V\propto T[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 456 ml  = 0.456 L

conversion used : (1 L = 1000 ml)

[tex]V_2[/tex] = final volume of gas = 3.4 L

[tex]T_1[/tex] = initial temperature of gas = ?

[tex]T_2[/tex] = final temperature of gas = [tex]65^oC=273+65=338K[/tex]

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

[tex]\frac{0.456L}{3.4L}=\frac{T_1}{338K}[/tex]

[tex]T_1=45.33K[/tex]

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

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