Question

One mole of ammonia gas is contained in a vessel which is capable of changing its volume (a compartment sealed by a piston, for example). The total energy U (in Joules) of the ammonia is a function of the volume V (in cubic meters) of the container, and the temperature T (in degrees Kelvin) of the gas. The differential dU is given by dU=840dV+27.32dT. (a) How does the energy change if the volume is held constant and the temperature is decreased slightly?

Answer 1

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!!!