A fish at a pressure of 1.1 atm has its swim bladder inflated to an initial volume of 8.16 mL. If the fish starts swimming horizontally, its temperature increases from 20.0 degrees Celsius to 22.0 degress Celcius as a result of exertion. A)Since the fish is still at the same pressure, how much work is done by the air in the swim bladder B)How much heat is gained by the air in the swim bladder? (Assume air to be a diatomic ideal gas) C)If this quantity of heat is lost by the fish, by how much will its temperature decrease? The fish has a mass of 5.00g and its specific heat is 3.5 J/(g*C)
To find the answers to these questions, we need to use the ideal gas law and the concept of work done by a gas, as well as the concept of heat gained or lost by an object.
Let's address each part of the question step by step:
A) To find the work done by the air in the swim bladder, we need to calculate the change in volume of the swim bladder. We can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the initial and final temperatures from Celsius to Kelvin:
Initial temperature (T1) = 20.0 degrees Celsius + 273.15 = 293.15 K
Final temperature (T2) = 22.0 degrees Celsius + 273.15 = 295.15 K
Since the fish is still at the same pressure (1.1 atm), we can write the equation as:
P * V1 / T1 = n * R
P * V2 / T2 = n * R
Since n and R are constant, we can equate the two equations:
P * V1 / T1 = P * V2 / T2
Rearranging the equation, we can solve for the change in volume (ΔV = V2 - V1):
ΔV = (P * V2 * T1 - P * V1 * T2) / (P * T1)
Plugging in the given values:
P = 1.1 atm
V1 = 8.16 mL = 8.16 cm³
V2 = ?
T1 = 293.15 K
T2 = 295.15 K
Converting the volumes to liters (1 cm³ = 1 mL) and the pressure to Pascals (1 atm = 101325 Pa), we get:
ΔV = (1.1 * 8.16 * 10^(-3) * 293.15 - 1.1 * 8.16 * 10^(-3) * 295.15) / (1.1 * 293.15)
Simplifying the equation, we find:
ΔV ≈ -0.0055 L
Since the volume change is negative, the work done by the air in the swim bladder is positive. The work done can be calculated using the equation:
work = -P * ΔV
Plugging in the values:
work = -(1.1 atm * (-0.0055 L))
work ≈ 0.006 L·atm
So, the work done by the air in the swim bladder is approximately 0.006 L·atm.
B) To find the heat gained by the air in the swim bladder, we can use the equation:
q = n * C * ΔT
where q is the heat gained or lost, n is the number of moles of gas, C is the molar specific heat capacity of the gas, and ΔT is the change in temperature.
Since air is assumed to be a diatomic ideal gas, its molar specific heat capacity at constant volume (Cv) is 20.8 J/(mol·K), and we can calculate the number of moles using the ideal gas law:
PV / (RT) = n
Plugging in the values:
P = 1.1 atm
V = 8.16 mL = 8.16 cm³
R = 0.0821 L·atm/(mol·K)
T = 293.15 K
n = (1.1 * 8.16 * 10^(-3)) / (0.0821 * 293.15)
Simplifying the equation, we find:
n ≈ 0.000383 mol
Now we can calculate the heat gained:
q = 0.000383 mol * 20.8 J/(mol·K) * (295.15 K - 293.15 K)
Simplifying the equation, we find:
q ≈ 0.0157 J
So, the air in the swim bladder gains approximately 0.0157 J of heat.
C) Now let's calculate the decrease in temperature of the fish using the concept of heat gained or lost by an object. The heat lost by the fish can be calculated using the equation:
q = m * C * ΔT
where q is the heat gained or lost, m is the mass of the fish, C is the specific heat capacity of the fish, and ΔT is the change in temperature.
Plugging in the values:
m = 5.00 g
C = 3.5 J/(g·K)
q = 0.0157 J
We can solve the equation for ΔT:
ΔT = q / (m * C)
Substituting the values, we find:
ΔT = 0.0157 J / (5.00 g * 3.5 J/(g·K))
Simplifying the equation, we find:
ΔT ≈ 0.000894 K
So, the temperature of the fish would decrease by approximately 0.000894 degrees Celsius.