Wize University Physics Textbook (Master) > Thermodynamics

Ideal Gases and First Law of Thermodynamics

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Ideal Gas Law

For any ideal gas, the following equation shows the relation between its thermodynamic variables:
PV=nRT=NkBTPV=nRT=Nk_B T
  • PP is the pressure of the gas in PaPa
  • VV is the volume of the gas in m3m^3
  • nn is the number of moles of the gas
  • RR is the gas constant and is equal to R=8.31446JmolKR=8.31446 \frac{J}{mol\cdot K}
  • NN is the number of atoms in the gas
  • kBk_B is the Boltzmann constant which is equal to:
kB=1.38×1023J/Kk_B=1.38\times10^{-23} J/K
  • kB=RNAk_B=\frac{R}{N_A} where NAN_A is the Avogadro constant
  • NAN_A is the number of atoms in one mole of a given substance and is equal to NA=6.022×10231molN_A=6.022\times10^{23} \frac{1}{mol}
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Example: Ideal Gases

A piston containing oxygen gas starts with state variables P1, V1, N, and T1. The piston expands slowly until it reaches a final volume V2 = 4V1 while maintaining the same temperature. Immediately after, the gas rapidly increases to a temperature of T3 = 1.1T1 but is no longer able to change in volume. What is the final pressure, P3?

Process 1 to 2: Isothermal : ΔT=T2T1=0       p2=?V2=4V1\Delta T=T_2-T_1=0\ \ \ \ \ \ \ p_2=?V_2=4V_1
Process 2 to 3: Isochoric : T3=1.11    p3=?          v=v2=4v1T_3=1.11\ \ \ \ p_3=?\ \ \ \ \ \ \ \ \ \ v_=v_2=4v_1
To find p2 : p2v2=nRT2nRT1=p1v1p_2v_2=nRT_2-nRT_1=p_1v_1
p2=p1v1v2=14p1\therefore p_2= p_1\frac{v_1}{v_2}=\frac{1}{4}p_1
To find p3 : p3v3=nRT3                p3=nRT3v3=1.14×nRT1v1=1.14p1=0.275p1p_3v_3=nRT_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p_3=\frac{nRT_3}{v_3}=\frac{1.1}{4}\times\frac{nRT_1}{v_1}=\frac{1.1}{4}p_1=0.275p_1

An ideal gas at 30.0 C and pressure of 2 × 105 Pa is in a container having a volume of 1.40 L.
  1. Determine the number of moles of gas in the container.
  2. The gas pushes against a piston, expanding to twice its original volume, while the pressure falls to atmospheric pressure. Find the final temperature.
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First Law of Thermodynamics

  • Change in the internal thermal energy of the system is either due to heat exchange or work done on the gas or both:
ΔU=QW\Delta U=Q-W
  • The internal thermal energy of an ideal gas is ONLY function of its temperature.
Important Note: How to find the sign of ∆U?


  • ΔU>0\Delta U>0 if ΔT>0\Delta T>0 and ΔU<0\Delta U<0 if T<0∆T<0
  • Q>0Q>0 if the heat is added to the gas
  • Q<0Q<0 if the heat is removed from the gas
  • W>0W>0 if the work is done by the gas
  • W<0W<0 if the work is done on the gas

Extra Practice