Question

Three gas samples are held at constant volume within their individual, separate containers. Container A has 8.5 moles of a dilute monatomic gas, container B has 7.6 moles of a dilute diatomic gas, and container C has 11.75 moles of a dilute polyatomic gas. Each gas receives the heat required for a temperature increase of 12.9K. What is the heat absorbed, in joules, by the gas in container A? What is the heat absorbed, in joules, by the gas in container B?

Answer 1

The heat absorbed by the gas in container A is 414.7 J, and the heat absorbed by the gas in container B is 640.9 J.

Step-by-step explanation:

To determine the heat absorbed by the gas in container A, we need to first calculate the change in internal energy of the gas, which can be done using the equation:

ΔU = nC_VΔT

Where ΔU is the change in internal energy, n is the number of moles of gas, C_V is the molar heat capacity at constant volume, and ΔT is the change in temperature.

For a monatomic gas, the molar heat capacity at constant volume (C_V) is approximately 3/2R, where R is the ideal gas constant.

So, in container A, with 8.5 moles of monatomic gas and a temperature increase of 12.9K, we can calculate the change in internal energy:

ΔU = 8.5 * (3/2 * 8.314 J/mol K) * 12.9 K

ΔU = 8.5 * 3/2 * 8.314 * 12.9 J

ΔU = 414.7 J

Therefore, the heat absorbed by the gas in container A is 414.7 J.

To determine the heat absorbed by the gas in container B, we use the same equation:

ΔU = nC_VΔT

For a diatomic gas, the molar heat capacity at constant volume (C_V) is approximately 5/2R.

So, in container B, with 7.6 moles of diatomic gas and a temperature increase of 12.9K, we can calculate the change in internal energy:

ΔU = 7.6 * (5/2 * 8.314 J/mol K) * 12.9 K

ΔU = 7.6 * 5/2 * 8.314 * 12.9 J

ΔU = 640.9 J

Therefore, the heat absorbed by the gas in container B is 640.9 J.