Question

A steel container of mass 135 g contains 24.0 g of ammonia, NH₃, which has a molar mass of 17.0 g/mol. The container and gas are in equilibrium at 12.0°C. How much heat has to be removed to reach a temperature of −20.0°C? Ignore the change in volume of the steel and treat the ammonia as an ideal gas with 6 degrees of freedom (3 translational and 3 rotational; at room temperature, the vibrational degree of freedom is inactive.)

(a) 640 J
(b) 860 J
(c) 1240 J
(d) 1490 J

Answer 1

To calculate the amount of heat that needs to be removed to reach -20.0°C, we need to consider the heat capacity of the container and the heat transferred from the ammonia gas. The heat transferred from the container can be calculated using the formula q = mcΔT.

Step-by-step explanation:

To calculate the amount of heat that needs to be removed to reach -20.0°C, we need to consider the heat capacity of the container and the heat transferred from the ammonia gas.

The heat capacity of the container can be calculated using the formula:

q = mcΔT

Where:

• q is the heat transferred
• m is the mass of the container
• c is the specific heat capacity of steel (0.46 J/g°C)
• ΔT is the change in temperature from 12.0°C to -20.0°C (-32.0°C)

Using the given mass of the container (135 g) and the specific heat capacity of steel, we can calculate the heat transferred from the container.

The heat transferred from the ammonia gas can be calculated using the formula:

q = nΔH

Where:

• q is the heat transferred
• n is the number of moles of ammonia gas
• ΔH is the enthalpy of vaporization of ammonia gas (4.8 kJ/mol)

Using the given mass of ammonia (24.0 g) and its molar mass (17.0 g/mol), we can calculate the number of moles of ammonia and then the heat transferred from the gas.

Finally, we can add the heat transferred from the container and the gas to get the total amount of heat that needs to be removed.

Therefore, the correct answer is (d) 1490 J.