Ideal Gas Law

The Essentials

The majority of materials that are analyzed in thermodynamics are fluids, which includes liquids and glasses. Liquids are generally assumed to be incompressible(volume does not change due to added pressure), but gasses are a compressible fluid. For this reason, a new equation is needed to relate the pressure, volume and temperature of a gas. For ideal gasses (gasses where pressure is low and temperature is much higher than the critical point), the following law applies, where $R$ is a gas constant that is given for the material. $$Pv=RT, PV=mRT$$ This equation can be used to relate two different states of an ideal gas. If there is no mass or substance change, the following relation can be used. $$(\frac{Pv}{RT})_1=(\frac{Pv}{RT})_2$$

Specific Heat Assumptions

For ideal gasses, internal energy is a function of only temperature, so the internal energy and enthalpy can be estimated using the change in temperature and the specific heat. The specific heat is the amount of heat needed to raise one unit mass by one deg C. There are two different specific heats used in thermodynamics, constant volume specific heat ($c_v$) and constant pressure specific heat($c_p$). These values are used in the following relationships to find an approximate change in internal energy and enthalpy. $$\Delta u=c_{v,ave}(T_2-T_1)$$ $$\Delta h=c_{p,ave}(T_2-T_1)$$ The "ave" in these equations stands for the average specific heat. The specific heat of a substance will change as the temperature changes, so using the average specific heat between the two states will result in a better estimate. This estimate is generally considered to be sufficiently accurate if the change in temperature is less than 100 deg C. Additionally, these approximations can be applied to liquids and solids, as long as the material is of one phase for the duration of the analysis. Note that for all ideal gas equations, temperature in Kelvin is used.

Example

1: Two tanks of air are connected by a valve. One of the tanks holds 2 kg of air at 77 deg C and 70 kPa. The other tank holds 8 kg of air at 27 deg C and 120 kPa. The valve is opened and the air mixes. At the same time, some heat is added to the system. The final temperature of the tanks is 42 deg C. Find the final pressure of the system, the heat added using constant specific heat assumptions, and the heat added using the Ideal Gas tables.

Air at these temperatures and pressures can be treated as an ideal gas, so the ideal equation can be applied to each individual state. It is useful to consider 3 states, the two separate tanks at the beginning and the final combined state at the end. The ideal gas law can be rewritten to solve for the final pressure as follows. $$P_3=\frac{m_3RT_3}{V_3}$$ The mass of state three will be the masses of the two tanks combined. The volume of the final state will be the sum of the volumes of the tanks, which can be found using the ideal gas law twice more. $$m_3=m_1+m_2=10\text{ kg}$$ $$V_3=V_2+V_1, V_1=\frac{m_1RT_1}{P_1},V_2=\frac{m_2RT_2}{P_2}$$ After given numbers are entered, the final pressure can be calculated. $$V_1=2.87\text{ m}^3, V_2=5.74\text{ m}^3, V_3=8.61\text{ m}^3$$ $$P_3=\frac{10*0.287*(42+273.15)}{8.61}=105.0\text{ kPa}$$ To find the heat added, start with the First Law and cancel out any terms that are known not to change. $$Q_{in}-\cancel{Q_{out}}+\cancel{W_{in}}-\cancel{W_{out}}=\Delta U+\cancel{\Delta KE}+\cancel{\Delta PE}$$ $$Q_{in}=m_1(u_3-u_1)+m_2(u_3-u_2)$$ First, solve for the heat transfer using the constant specific heat assumptions. Because the heat transfer varies only based on the specific internal energy, substitute the specific heat assumption into the simplified first law. $$\Delta u=c_v\Delta T\rightarrow Q_{in}=m_1c_{v_1}(T_3-T_1)+m_2c_{v_2}(T_3-T_2)$$ To find the specific heat, reference table A-2. Use the average specific heat value between the two temperatures. $$c_{v_1}=\frac{c_{v@350K}+c_{v@315K}}{2}=0.720\text{ kJ/kg K}$$ $$c_{v_2}=\frac{c_{v@350K}+c_{v@300K}}{2}=0.719\text{ kJ/kg K}$$ Enter values into the equation above and solve. $$Q_{in}=2*0.72*(315-350)+8*.719*(315-300)=35.76\text{ kJ}$$ Next, solve for the heat transfer using the exact ideal gas tables. The same simplified form of the First Law can be used, meaning that only the specific internal energy values are needed. Reference these values using the temperature of each state. $$Q_{in}=2*(224.85-250.02)+8*(224.85-214.07)=35.9\text{ kJ}$$

Practice

1: A airship is at 30 kPa and 20 deg C while anchored on the ground. The airship rises until the temperature is -10 deg C. What is the pressure inside the airship?

Solutions:

1: 26.93 kPa

More Resources

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