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3 years ago

How is specific heat capacity related to a change in state of matter?

1 answer:

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The heat capacity or thermal capacity of a body is the quotient between the amount of heat energy transferred to a body or system in any process and the change in temperature it experiences. In a more rigorous form, it is the energy necessary to increase the temperature of a certain substance by one temperature unit. [1] It indicates the greater or lesser difficulty that said body presents in experiencing changes in temperature under the supply of heat. It can be interpreted as a measure of thermal inertia. It is an extensive property, since its magnitude depends not only on the substance but also on the amount of matter in the body or system; therefore, it is characteristic of a particular body or system. For example, the heat capacity of the water in an Olympic-size swimming pool will be greater than that of the water in a glass. In general, heat capacity also depends on temperature and pressure.

Explanation:

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1) List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

2.1- List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

2.2- Convert grams of O2 to moles of O2.

The molar mass of O2 is 31.9988 g/mol.

$mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2$

2.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

2.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}$ $P_{O_2}=1.57\text{ }atm$

The pressure of O2 is 1.57 atm.

3) Find the pressure of F2.

3.1- List the known and unknown quantities.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- Convert grams of F2 to moles of F2.

The mmolar mass of F2 is 37.9968 g/mol.

$mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2$

3.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

3.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}$ $P_{F_2}=1.13\text{ }atm$

The pressure of F2 is 1.13 atm.

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

4.1- Set the equation.

$P_T=P_A+P_B$

4.2- Plug in the known quantities.

$P_T=1.57\text{ }atm+1.13\text{ }atm$ $P_T=2.7\text{ }atm$

The total pressure in the container is 2.7 atm.

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10 months ago