Skip to content
III. Long Answer Questions · Q2

Q.Discuss the ideal gas laws.

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★est
6% · 7/126 Questions
✓ Free question

Step 1. Boyle's law (constant temperature): the pressure of a fixed amount of gas is inversely proportional to its volume, P∝1VP\propto\dfrac{1}{V}.

Step 2. Charles' law (constant pressure): the volume of a fixed amount of gas is directly proportional to its absolute temperature, V∝TV\propto T.

Step 3. Combining the two gives PV=CTPV=CT for some constant CC. Considering two identical gas containers (same P,V,T,NP,V,T,N) merged into a single combined system -- now with volume 2V2V and particle count 2N2N, but unchanged P,TP,T -- shows that CC must scale with the particle number NN: C=NkC=Nk, where k=1.381×10−23 J K−1k=1.381\times10^{-23}\ \text{J K}^{-1} is the universal Boltzmann constant.

Step 4. This gives the ideal gas law PV=NkTPV=NkT. Writing N=μNAN=\mu N_A (Avogadro's number NA=6.023×1023 mol−1N_A=6.023\times10^{23}\ \text{mol}^{-1}) and R=NAk=8.314 J mol−1K−1R=N_Ak=8.314\ \text{J mol}^{-1}\text{K}^{-1} (universal gas constant) turns this into the molar form PV=μRTPV=\mu RT -- the standard equation of state for an ideal gas, valid at thermodynamic equilibrium. One mole occupies 22.4 L at STP and about 24.6 L at 300 K.

✓Final answer

The ideal gas law, PV=NkTPV=NkT (equivalently PV=μRTPV=\mu RT), follows from combining Boyle's law (P∝1/VP\propto1/V) and Charles' law (V∝TV\propto T) with the particle-scaling argument C=NkC=Nk.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.