Collection of Solved Problems in Physics

Enthalpy Change of Nitrogen

Task number: 3942

• Given Values

  m = 2 g = 0.002 kg	nitrogen mass
  t1 = 27 °C => T1 = 300.15 K	initial temperature of nitrogen
  V2 = 3/4 V1	final volume of nitrogen
  ΔH = ?	enthalpy change of nitrogen

  Table values:

  cp = 1042 J kg−1 K−1

  specific heat capacity of nitrogen

• Hint 1 – Enthalpy H

  Enthalpy H of gas is a thermodynamic potential defined by the relation

  \[H=U+pV,\]

  where U is inner energy, p pressure and V volume of the gas.

  The unit of enthalpy is a joule.

• Hint 2 – Elementary Change of Enthalpy dH

  The elementary change of enthalpy dH can be determined by differentiation of defining relation for enthalpy, therefore

  \[\text{d}H = \text{d}U+p\,\text{d}V+V\,\text{d}p.\]

  However, we do not know the values of some variables in this equation. How can we adjust this equation, so that we would get rid of unknown elementary change of inner energy on the right side?

• Hint Solution

  Using a mathematical expression of the Second Law of Thermodynamics in the form of

  \[\text{d}U=T\,\text{d}S-p\,\text{d}V,\]

  we obtain

  \[\text{d}H=\text{d}U+p\,\text{d}V+V\,\text{d}p = T\,\text{d}S+V\,\text{d}p.\]

  Now we have to realize that during isobaric process the change in pressure is equal to zero, i.e. dp = 0. We can then write

  \[\text{d}H= T\,\text{d}S.\]

• Hint 3 – Elementary Entropy Change dS

  Elementary entropy change dS is defined by the relation

  \[\text{d}S=\frac{\text{d}Q}{T},\]

  where Q is amount of heat supplied to gas and T gas temperature.

• Hint 4 – Expressing Supplied Heat

  To express the supplied heat use the relation between heat Q and specific heat capacity c_p at constant pressure

  \[Q=nc_p\left(T_2-T_1\right),\]

  where n is the amount of substance, T₂ and T₁ are final and intial temperatures.

• Hint 5 – Temperature Change

  To express a change in temperature use Gay-Lussac's Law.

• Gay-Lussac's Law

  During isobaric process of ideal gas with constant mass, the volume of gas V is directly proportional to thermodynamic temperature T, ergo:

  \[\frac{V}{T}= const.\]

  or

  \[\frac{V_1}{T_1}=\frac{V_2}{T_2}.\]

• Analysis

  We start with a definition expression of enthalpy that we differentiate. Thus we obtain an elementary change of enthalpy. In order to express it using only given values, we use a mathematical expression of the Second Law of Thermodynamics. Furthermore we must realize that the change in pressure during isobaric process is zero.

  If we express the elementary change of entropy as a ratio of elementary change of heat and temperature, we realize that the total change of enthalpy is equal to heat supplied to gas. This can be expressed as a product of specific heat capacity at constant pressure, mass and temperature change of the gas.

  To determine temperature change of the gas we use Gay-Lusssac's Law.

• Solution

  Enthalpy H is a thermodynamic potential defined by expression

  \[H=U+pV,\]

  where U is internal energy, p pressure and V volume of gas.

  By differentiating the definition expression we obtain elementary change of enthalpy dH:

  \[\text{d}H=\text{d}U+p\,\text{d}V+V\,\text{d}p.\]

  If we use a mathematical formulation of the Second Law of Thermodynamics in the form of

  \[\text{d}U=T\,\text{d}S-p\,\text{d}V,\]

  we can write

  \[\text{d}H=\text{d}U+p\,\text{d}V+V\,\text{d}p = T\,\text{d}S+V\,\text{d}p.\]

  Due to the fact that pressure is constant during the process given in the assignment, its elementary change is equal to zero, therefore dp = 0, and we obtain

  \[\text{d}H=T\,\text{d}S.\]

  Using the definition relation for elementary change of entropy

  \[\text{d}S=\frac{\text{d}Q}{T}\]

  we can write

  \[\text{d}H=T\,\text{d}S= T\frac{\text{d}Q}{T}= \text{d}Q.\]

  After the integration of previous relation we obtain for the total change of enthalpy

  \[\Delta H=Q,\]

  i.e. enthalpy change during isobaric process corresponds to the amount of heat supplied to the gas.

  For enthalpy change during isobaric compression therefore applies the following

  \[\Delta H=Q=c_pm\left(T_2-T_1\right),\]

  where c_p is specific heat capacity of gas, m mass of gas, T₁ and T₂ temperature of gas before and after the compression.

  During isobaric process with ideal gas with constant mass, the volume of the gas V is directly proportional to thermodynamic temperature T. If the volume decreases by one quarter, the temperature also decreases by one quarter.

  After substituting for T₂ = 3T₁/4 we obtain

  \[\Delta H= c_pm\left(\frac{3}{4}T_1-T_1\right)= -\frac{1}{4}c_pmT_1.\]

  The minus sign means that the heat was not supplied to the gas, it was given off by the gas (which corresponds to isobaric compression), and enthaply therefore decreases during this process.

• Numerical Solution

  \[\Delta H= -\frac{1}{4}c_pmT_1=-\frac{1}{4}\cdot 1042\cdot{ 0.002}\cdot{ 300.15}\,\mathrm{J}\dot{=} -160\,\mathrm{J}\]

• Answer

  Gas enthalpy during isobaric compression decreased by approximately 160 J.