Question

What Celsius temperature, T_2, is required to change the volume of the gas sample in Part A (T_1 = 12 degree C, V_1 = 1 77 times 10^3 L) to a volume of 3.54 times 10^3 L? Assume no change in pressure or the amount of gas in the balloon. Express your answer with the appropriate units.

Answer 1

Ans. Charles’ Law: Volume (V) of a gas in a closed vessel is directly proportional to absolute temperature (T) at constant pressure. V/T for a gas is a constant at specified pressure.

                                    (V1/ T1) = (V2/ T2)            - equation 1

Where, V1 = 1.77 x10³ L at T1=12⁰C = 285.15 K

V2 = 3.54 x10³ L at T2

Putting the values in equation 1-

            (1.77 x 10³ L) / 285.15 K = (3.54 x 10³ L) / T2

            Or, T2 = (3.54 x 10³ L) / [(1.77 x 10³ L) / 285.15 K] = 570.3 K

Therefore, T2 = 570.3 K = 297.15⁰C = 297⁰C