Answer:
a) Option C is correct.
It decreases slightly.
b) Option A is correct.
It increases slightly.
c) dU = 108.986 J = 109 J
Step-by-step explanation:
dU = 840dV + 27.32dT
With U = Total Energy of Ammonia in Joules
V = volume of ammonia (in cubic meters)
T = Temperature of Ammonia in Kelvin.
a) How does the energy change if the volume is held constant and the temperature is decreased slightly?
dU = 840dV + 27.32dT
If the volume is held constant, dV = 0
Then the temperature is decreased slightly, dT = -x (where x is a small number; minus sign indicates a decrease in temperature)
dU = 840(0) + 27.32 (-x)
dU = -27.32x
Obviously a slight decrease in total energy; due to the minus sign. A negative change indicates decrease.
b) How does the energy change if the temperature is held constant and the volume is increased slightly?
dU = 840dV + 27.32dT
If the temperature is held constant, dT = 0
Then the volume is increased slightly, dV = x (where x is a very small number)
dU = 840(x) + 27.32 (0)
dU = 840x
Obviously a slight increase in total energy; due to the positive sign. A positive change indicates increase.
c) Find the approximate change in energy if the gas is compressed by 350 cubic centimeters and heated by 4 degrees Kelvin.
The ammonia gas is compressed by 350 cm³
In the units for the equation,
350 cm³ = (350 × 10⁻⁶) m³ = (3.50 × 10⁻⁴) m³
So, the ammonia gas is compressed by (3.50 × 10⁻⁴) m³. A compression is a decrease in volume, hence,
dV = - (3.50 × 10⁻⁴) m³
The ammonia gas is heated by 4 K.
Heating indicates an increase in temperature, hence,
dT = 4 K
dU = 840dV + 27.32dT
dU = 840 (-3.50 × 10⁻⁴) + 27.32 (4)
dU = -0.294 + 109.28 = 108.986 J ≈ 109 J
Hope this Helps!!!
T = r
