Heat capacities Cp, Cv; relationship Cp − Cv = nR (ideal gas)

Derivation of Cp − Cv = nR (Ideal Gas)

Step 1: First Law Foundation

Start with the first law of thermodynamics: $dU = \delta Q - \delta W = \delta Q - P\,dV$

For an ideal gas, internal energy depends only on temperature: $dU = nC_v\,dT$

WHY? For ideal gases, molecules don't interact (no potential energy between them), so $U$ depends only on kinetic energy → only $T$ .

Step 2: Enthalpy Definition

Enthalpy is defined as: $H = U + PV$

Take the differential: $dH = dU + P\,dV + V\,dP$

Substitute $dU = nC_v\,dT$ and rearrange: $dH = nC_v\,dT + P\,dV + V\,dP$

Step 3: Apply Constant Pressure Condition

At constant pressure ( $dP = 0$ ): $dH = nC_v\,dT + P\,dV$

But by definition, at constant pressure: $dH = nC_p\,dT$

Equating these: $nC_p\,dT = nC_v\,dT + P\,dV$

Step 4: Use Ideal Gas Law

For an ideal gas at constant pressure, differentiate $PV = nRT$ : $P\,dV = nR\,dT$

WHY? Since $P$ is constant, $d(PV) = P\,dV = d(nRT) = nR\,dT$ .

Step 5: Substitute and Simplify

Replace $P\,dV$ in Step 3: $nC_p\,dT = nC_v\,dT + nR\,dT$

Divide through by $n\,dT$ :

Physical Interpretation

The nR term represents the work done by the gas during expansion when heated at constant pressure.

At constant V: No expansion → $Q = \Delta U = nC_v\Delta T$
At constant P: Expansion happens → $Q = \Delta H = nC_p\Delta T =_v\Delta T + nR\Delta T$

The extra $nR\Delta T$ is the $P\Delta V$ work energy that "leaks out" pushing against external pressure.

Worked example Worked Example 2: Heat Required at Different Conditions

Given: 2 moles of nitrogen gas (N₂, diatomic) with $C_v = \frac{5}{2}R$ . Temperature increases from 300 K to 350 K.

Find: Heat required (a) at constant volume, (b) at constant pressure.

Solution (a): At constant V: $Q_v = nC_v\Delta T = 2 \times \frac{5}{2}(8.314) \times 50$ $Q_v = 2077\,\text{J}$

Why this formula? No work done, so all heat becomes internal energy: $Q = \Delta U = nC_v\Delta T$ .

Solution (b): At constant P, first find $C_p$ : $C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$ $Q_p = nC_p\Delta T = 2 \times \frac{7}{2}(8.314) \times 50$ $Q_p = 2910\,\text{J}$

Why more heat? The difference $Q_p - Q_v = 2nR\Delta T = 2(2)(8.314)(50) = 831\,\text{J}$ is the expansion work done.

Verify work: $W = P\Delta V = nR\Delta T = 2(8.314)(50) = 831\,\text{J}$ ✓

Heat Capacity Ratio γ (Gamma)

The adiabatic index or heat capacity ratio: $\gamma = \frac{C_p}{C_v} = \frac{C_v + R}{C_v} = 1 + \frac{R}{C_v}$

Values for ideal gases:

Monatomic (He, Ar): $\gamma = 5/3 ≈ 1.67$
Diatomic (N₂, O₂): $\gamma = 7/5 = 1.40$
Polyatomic (CO₂, CH₄): $\gamma ≈ 1.29 - 1.33$

WHY does γ decrease? More degrees of freedom (rotational, vibrational) increase $C_v$ , so the ratio $C_p/C_v$ decreases.

Memory Aids

Recall Feynman Technique: Explain to a 12-Year-Old

Imagine you have a balloon filled with air. You want to make the air inside hotter.

Method 1: Hold the balloon tight so it can't expand. When you add heat, all the energy goes into making the air molecules move faster (getting hotter).

Method 2: Let the balloon expand freely while you heat it. Now some of your heat energy is "wasted" pushing the balloon bigger instead of just making molecules move faster. So you need MORE heat to get the same temperature increase!

The "extra heat" needed in Method 2 is always the same amount per molecule: it's called $R$ (the gas constant). That's why scientists say $C_p - C_v = R$ . It's the "expansion tax"—the price you pay for letting the gas grow while heating it!

Connections

First Law of Thermodynamics — Foundation for deriving $C_p - C_v$
Enthalpy — Why $C_p$ involves enthalpy; relationship $H = U + PV$
Ideal Gas Law — Used to relate $P\,dV$ to $nR\,dT$
Adiabatic Process — Uses $\gamma = C_p/C_v$ in equation $PV^\gamma = \text{const}$
Degrees of Freedom — Determines $C_v$ values (equipartition theorem)
Internal Energy — For ideal gases, $dU = nC_v\,dT$ always
Isothermal vs Adiabatic Processes — Heat capacities determine process behavior
Real Gases (van der Waals) — Deviations from $C_p - C_v = R$

#flashcards/chemistry

What is Cv (molar heat capacity at constant volume)? :: The heat required to raise 1 mole of substance by 1 K at constant volume; equals (∂U/∂T)_V. For ideal gases, all heat becomes internal energy since no work is done.

What is Cp (molar heat capacity at constant pressure)? :: The heat required to raise 1 mole of substance by 1 K at constant pressure; equals (∂H/∂T)_P. More heat is needed than Cv because the gas does expansion work.

State Mayer's relation for ideal gases.

Cp − Cv = R (for 1 mole) or Cp − Cv = nR (for n moles). The difference R represents the expansion work per mole per kelvin.

Why is Cp always greater than Cv for gases?

At constant pressure, the gas expands during heating, doing work W = PΔV = nRΔT. This work energy is "extra" beyond the internal energy increase, so more heat is needed: Cp = Cv + R.

For an ideal gas, what isΔU in terms of Cv?

ΔU = nCvΔT, valid for ANY process (not just constant volume), because U depends only on T for ideal gases.

What is the heat capacity ratio γ?

γ = Cp/Cv; for monatomic gases γ = 5/3, for diatomic γ = 7/5. Used in adiabatic process equations: PV^γ = constant.

Is Cp − Cv = R valid for liquids and solids?

No, it's exact only for ideal gases. For condensed phases, Cp − Cv = TVα²/κ_T (usually negligible), so Cp ≈ Cv.

For 2 moles of He (Cv = 3R/2), what is the heat at constant P forΔT = 50 K?

First find Cp = Cv + R = 5R/2. Then Q = nCpΔT = 2(5R/2)(50) = 250R = 2078 J.

What does the nRΔT term represent physically in Qp − Qv?

The expansion work W = PΔV = nRΔT done by the gas when heated at constant pressure.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Heat Capacity kya hai aur Cp-Cv ka relation kyun important hai?

Jab hum kisi gas ko garam karte hain, to heat capacity bati hai kitna heat chahiye 1 degree temperature badhane ke liye. Lekin ek twist hai—agar gas ko constant volume pe (ek sealed rigid container mein) heat karo, to ek value milti hai jise Cv kehte hain. Agar gas ko constant pressure pe (jahan gas expand kar sakta hai, jaise piston-cylinder mein) heat karo, to zyada heat lagti hai, use Cp kehte hain. Yeh difference kyun? Kyunki constant pressure mein gas expand hota hai, to kuch energy expansion work mein chali jaati hai—piston ko push karne mein.

Formula hai: Cp − Cv = R (jahan R universal gas constant hai, 8.314 J/mol·K). Iska physical matlab hai ki jo extra heat chahiye constant pressure mein, wo exactly expansion work ke barabar hai. Ideal gas ke liye yeh relation bilkul exact hai. Real gases ya liquids/solids ke liye thoda complicated ho jata hai. Yeh relation thermodynamics ke sabse elegant results mein se ek hai—first law aur ideal gas law se directly derive hota hai. Jab bhi adiabatic processes (jahan heat exchange nahi hota) solve karte hain, tab γ = Cp/Cv ratio use hota hai. Monatomic gas (He, Ar) ke liye γ = 5/3, diatomic (N₂, O₂) ke liye 7/5. Yeh difference samajhna zaroori hai taki hum predict kar sakein ki gas kis tarah behave karega different heating conditions mein—chemistry aur physics dono mein bohot fundamental concept hai.

Heat capacities Cp, Cv; relationship Cp − Cv = nR (ideal gas)

Core Definitions

Derivation of Cp − Cv = nR (Ideal Gas)

Step 1: First Law Foundation

Step 2: Enthalpy Definition

Step 3: Apply Constant Pressure Condition

Step 4: Use Ideal Gas Law

Step 5: Substitute and Simplify

Physical Interpretation

Heat Capacity Ratio γ (Gamma)

Common Mistakes

Memory Aids

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Thermodynamics (Chemical)

Connections