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kondor19780726 [428]
3 years ago
11

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

Chemistry
2 answers:
alexira [117]3 years ago
5 0
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
ValentinkaMS [17]3 years ago
5 0

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

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