Helpp with thermodynamics and specific heat question

[TheTourist]

For an ideal monatomic gas, the molar specific heat at constant volume,
C_v=3/2R, where R is the gas constant. The heat capacity at constant volume of a gas is measured to be 43.58kJK^-1. How many kg-moles of the gas are present?




Iv tried looking in the lecture notes and in the textbook, but can't find anything of use.

Iv made an attempt and got 3.5kg-moles. Not sure that its right though. I used R=kN_A, where k is Boltzmann constant and N_A is Avogadros number.

 

[Andrew Mason]

For an ideal monatomic gas, the molar specific heat at constant volume,
C_v=3/2R, where R is the gas constant. The heat capacity at constant volume of a gas is measured to be 43.58kJK^-1. How many kg-moles of the gas are present?




Iv tried looking in the lecture notes and in the textbook, but can't find anything of use.

Iv made an attempt and got 3.5kg-moles. Not sure that its right though. I used R=kN_A, where k is Boltzmann constant and N_A is Avogadros number.

The heat capacity of n moles is n times the molar heat capacity.

The heat capacity of one mole of gas at constant volume is that amount heat flow required to raise the temperature one degree:

[tex]C_v = Q/\Delta T[/tex]

If there are n moles, multiply both sides by n to determine the amount of heat flow per degree of temperature change.

AM

 

[kerimek]

Hello,

Thank you for reaching out for help with your thermodynamics and specific heat question. The molar specific heat at constant volume, Cv, for an ideal monatomic gas is indeed 3/2R, where R is the gas constant. However, the heat capacity at constant volume, Cv, is measured in units of energy per temperature, not energy per temperature per mole. Therefore, to solve for the number of kg-moles of gas present, we need to use the following equation:

Cv = (3/2)R * n

Where n is the number of moles of gas present. To solve for n, we can rearrange the equation as follows:

n = (Cv * 2)/(3R)

Plugging in the given value for Cv (43.58 kJ/K) and the value for R (8.314 J/molK), we get:

n = (43.58 kJ/K * 2)/(3 * 8.314 J/molK) = 3.3 kg-moles

Therefore, there are approximately 3.3 kg-moles of gas present in this system. Your attempt of 3.5 kg-moles is quite close, but it is important to remember to use consistent units in your calculations. I hope this helps clarify the solution for you. Keep up the good work in your studies!