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(a) Compute the specific heat at constant volume of

Chapter 18, Problem 43E

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QUESTION:

(a) Compute the specific heat at constant volume of nitrogen (N2) gas, and compare it with the specific heat of liquid water. The molar mass of N2 is 28.0 g/mol. (b) You warm 1.00 kg of water at a constant volume of 1.00 L from 20.0o C to 30.0o C in a kettle. For the same amount of heat, how many kilograms of 20.0o C air would you be able to warm to 30.0o C? What volume (in liters) would this air occupy at 20.0o C and a pressure of 1.00 atm? Make the simplifying assumption that air is 100% N2.

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QUESTION:

(a) Compute the specific heat at constant volume of nitrogen (N2) gas, and compare it with the specific heat of liquid water. The molar mass of N2 is 28.0 g/mol. (b) You warm 1.00 kg of water at a constant volume of 1.00 L from 20.0o C to 30.0o C in a kettle. For the same amount of heat, how many kilograms of 20.0o C air would you be able to warm to 30.0o C? What volume (in liters) would this air occupy at 20.0o C and a pressure of 1.00 atm? Make the simplifying assumption that air is 100% N2.

ANSWER:

Solution 43E Step 1 If the temperature of a gas is T , then the energy per degrees of freedom is given by 1 q = k2TB Where k B is the boltzmann’s constant. Since Nitrogen (N 2is a diatomic molecule, the degrees of freedom of nitrogen molecule is 5. Hence the energy per molecule is q = 5 × ( k T) = k T 5 m 2 B 2 B 23 Now in one mole of N t2re will be N = 6.02 × 10 number of molecules, hence the total energy of the gas is 5 5 Q = N · ( 2 B) = Nk 2 B Now, suppose Q, heat is supplied to the system. Since the process is constant volume process, there will be no work done. Hence, so all energy will be used to change in temperature, hence the change in temperature is given by 5 Q = Nk2T B So the molar specific heat at constant volume is Q C v = Nk = (6.02 × 10 23 /mol)(1.38 × 10 23J/K) = 20.8 J/mol · K T 2 B 2 3 Now, the molar mass of N i2M = 28.0 g/mol = 28.0 × 10 kg/mol . Hence the specific at constant volume is given by Cv 20.8 J/mol·K c v M = 28.0×103kg/mol= 743 J/kg · K Now we know that the specific heat of water is c = 4190 J/kg · K w Which is much greater than the specific heat of nitrogen at constant volume.

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