A diving bell, containing at the initial moment of time v = 4 moles of air with a volume of V1 = 14 l

A diving bell, containing at the initial moment of time v = 4 moles of air with a volume of V1 = 14 l, is slowly lowered to the bottom of the reservoir. When this occurs isothermal compression of air to the final volume V2. The work done by water when compressing air is determined by the expression A = Alpha * v * Tlog2V1 / V2 (J), where (alpha = 11.6) is constant, and T = 300K is the air temperature. What volume of V2 (in liters) will the air occupy if a work of 27,840 J was performed during the compression of the gas?

First, we substitute the quantities we know into the logarithmic equation:
27840 = 11.6 * 4 * 300 * log2 (14 / V2).
The next step is to transform this logarithmic equation and find its value:
log2 (14 / V2) = 27840 / (11.6 * 4 * 300) = 2.
For what a logarithm is, let’s recall its definition:
The logarithm is the exponent to which the base must be raised to get the number under the logarithm.
Next, let’s solve the equation based on the above:
14 / V2 = 2 ^ 2 = 4.
V2 = 14/4 = 3.5.
Answer: 3.5.