Q.What amount of heat must be supplied to 2.0×10−2 kg of nitrogen (at room temperature) to raise its temperature by 45∘C at constant pressure? (Molecular mass of N2=28; R=8.3 J mol−1 K−1.)
Concept understanding — Heat Capacity at Constant Pressure
Heat Capacity at Constant Pressure — From Intuition to Precision
Imagine you have a pot of water on a stove. You turn the burner on, and the water gets hotter. How much heat does it take to raise its temperature by, say, 10°C? That depends on two things: how much water you have, and whether the pot is open to the air or sealed tight.
If the pot is open (constant pressure — the air above it is always at atmospheric pressure), the water can expand as it heats. Some of the energy you supply goes into pushing the atmosphere aside — doing work against the outside air. So you need to put in more heat than if the pot were sealed (constant volume), where no expansion work is possible.
That extra heat is the key idea behind heat capacity at constant pressure, denoted Cp.
The Intuition First
Heat capacity tells you: "How much heat must I add to raise the temperature of this substance by 1°C (or 1 K)?"
- At constant volume (Cv): All the heat goes into increasing the internal energy (the kinetic and potential energy of the molecules). No work is done because the volume doesn't change.
- At constant pressure (Cp): Some heat goes into internal energy, but some also goes into the work of expansion against the constant external pressure. So Cp is always larger than Cv for gases (and for most solids/liquids, the difference is tiny because they barely expand).
For an ideal gas, the difference is exactly Cp−Cv=nR, where n is the number of moles and R is the universal gas constant. This is a direct consequence of the first law of thermodynamics.
The Precise Statement
Heat capacity at constant pressure is defined as the amount of heat required to raise the temperature of a substance by 1 K (or 1°C) while keeping the pressure constant.
Mathematically:
Cp=(dTδQ)p
The subscript p means "at constant pressure." The δQ (not dQ) reminds us that heat is a path-dependent quantity, not a state function.
But we can rewrite this in terms of a state function — enthalpy (H). At constant pressure, the heat added equals the change in enthalpy:
δQp=dH
Therefore:
Cp=(∂T∂H)p
This is the working definition you'll use in problems: Cp is the partial derivative of enthalpy with respect to temperature at constant pressure.
Molar vs. Specific Heat Capacity
You'll encounter two common forms:
- Molar heat capacity at constant pressure (Cp,m): heat capacity per mole (units: J mol⁻¹ K⁻¹)
- Specific heat capacity at constant pressure (cp): heat capacity per unit mass (units: J kg⁻¹ K⁻¹)
The total heat capacity of a sample is:
Cp=n⋅Cp,m=m⋅cp
Why It Matters
In most chemical reactions and physical processes, the system is open to the atmosphere — constant pressure. So Cp is the relevant quantity for:
- Calculating enthalpy changes (ΔH=nCp,mΔT)
- Designing calorimeters (like coffee-cup calorimeters that operate at constant pressure)
- Understanding why gases heat up when compressed (and cool when expanded)
Do not confuse Cp with Cv. For gases, the difference is significant. For solids and liquids, the difference is often negligible (typically less than 1%), so many textbooks treat them as approximately equal for condensed phases.
A Quick Example
How much heat is needed to raise the temperature of 2 moles of an ideal gas from 300 K to 400 K at constant pressure? (Given Cp,m=29.1 J mol−1K−1)
Qp=nCp,mΔT=(2)(29.1)(100)=5820 J
If the same gas were heated at constant volume, you'd need less heat — about 5820−nRΔT=5820−(2)(8.314)(100)=4157 J — because no expansion work is done.
Final takeaway: Cp is the heat capacity you measure when the system is free to expand against a constant external pressure. It's always larger than Cv for gases, and the difference comes from the work of expansion.
Students searching for "Heat Capacity at Constant Pressure: Definition, Formula & Real-World Examples" or "Heat Capacity at Constant Pressure 11 physics" will find this explanation directly aligned with the Class 11 Physics curriculum prescribed under NCERT/CBSE. It is also a recurring theme in JEE Main, NEET and state engineering/medical entrance exams, so working through it carefully pays off well beyond board exams.
Concept: Heat capacity at constant pressure for an ideal gas.
For a diatomic gas like nitrogen at room temperature, the molar heat capacity at constant pressure is Cp=27R. The heat required is given by:
Q=nCpΔT
where n is the number of moles.
Step 1: Calculate the number of moles.
n=Mm=28×10−3 kg mol−12.0×10−2 kg=2820=0.714 mol
Step 2: Substitute into the heat equation with ΔT=45 K (since a change in Celsius equals a change in Kelvin).
Q=0.714×27×8.3×45
Q=0.714×3.5×8.3×45=933 J
The heat required is 933 J or approximately 0.93 kJ.
For an ideal gas at constant pressure, heat supplied is Q=nCpΔT. Using Cp=27R for diatomic nitrogen, the heat required is 933J.
When we heat a gas at constant pressure, it not only gains internal energy but also does work by expanding against the external pressure. This is why the heat capacity at constant pressure, Cp, is always larger than at constant volume, Cv. For an ideal gas, the relationship is Cp=Cv+R.
Nitrogen is a diatomic molecule. At room temperature, it has three translational and two rotational degrees of freedom (vibrational modes are not excited). By the equipartition theorem, each degree of freedom contributes 21R per mole to the molar heat capacity at constant volume:
Cv=25R
Therefore, the molar heat capacity at constant pressure is:
Cp=Cv+R=25R+R=27R
Cp=27R=27×8.3=29.05J mol−1K−1
Now let's calculate the heat required step by step:
-
Find the number of moles of nitrogen.
Given mass m=2.0×10−2kg=20g and molecular mass M=28g mol−1:
n=Mm=2820=75mol
-
Identify the temperature change.
The temperature rise is ΔT=45∘C=45K (since a change in Celsius equals a change in Kelvin).
-
Apply the heat capacity formula at constant pressure.
The heat supplied at constant pressure is:
Q=nCpΔT
Substituting the values:
Q=75×27×8.3×45
-
Simplify the calculation.
Notice that 75×27=25:
Q=25×8.3×45=2.5×8.3×45
Q=2.5×373.5=933.75J
A common mistake is to use Cv instead of Cp when the problem specifies constant pressure. Always check whether the process is isobaric (constant P) or isochoric (constant V).
The amount of heat that must be supplied is 933J (or 934J if rounded).
A quick cross-check: nitrogen's tabulated specific heat at constant pressure is about cp≈1.04 J g−1K−1, so Q=mcpΔT=20 g×1.04×45≈936 J — matching the kinetic-theory answer to within rounding. This is a useful shortcut whenever you have a handy tabulated specific heat: it lets you skip converting mass to moles and multiplying by Cp=27R entirely, and it's also a good way to sanity-check that the molar route was set up correctly.
Showing the 12 most recent of 15 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The ratio of Molar specific Heats (Cp/Cv) of Monoatomic gas is :(a) 5/3(b) 5/7(c) 7/9(d) 7/5.
›Reveal solutionSolution
A monoatomic gas has 3 degrees of freedom, giving Cv = (3/2)R, Cp = (5/2)R, and hence Cp/Cv = 5/3.
By the law of equipartition of energy, each degree of freedom contributes (1/2)RT to the internal energy per mole of an ideal gas.
A monoatomic gas (e.g. He, Ar) has only 3 translational degrees of freedom (no rotational or vibrational contribution for a point-like molecule at ordinary temperatures).
Internal energy per mole: U = (3/2)RT
Cv = dU/dT = (3/2)R
Using Mayer's relation, Cp - Cv = R, so Cp = (3/2)R + R = (5/2)R
Cp/Cv = (5/2 R) / (3/2 R) = 5/3
✓Final answerThe correct option is (a) 5/3.
- CBSE 2026Set ANNUAL1 markMCQQ.The ratio of Cp and Cv for helium gas is(a) 5/3(b) 5/7(c) 3/5(d) 9/7
›Reveal solutionSolution
Helium is monatomic, so gamma = Cp/Cv = 5/3. Answer (A).
For a monatomic ideal gas (3 translational degrees of freedom):
Cv = (3/2)R and Cp = Cv + R = (5/2)R.
Therefore gamma = Cp/Cv = (5/2)R /((3/2)R) = 5/3.
Helium is monatomic, so its gamma is 5/3.
✓Final answer(A) 5/3.
- CBSE 2025Set ANNUAL1 markMCQQ.Mayer's formula is (A) C_P - C_V = R (B) C_P/C_V = R (C) C_V - C_P = R (D) C_V/C_P = R
›Reveal solutionSolution
Mayer's formula states CP−CV=R for one mole of an ideal gas.
At constant volume, all the heat supplied goes into raising internal energy: dQ=CVdT=dU.
At constant pressure, the gas also does work as it expands, so extra heat is needed for the same temperature rise: dQ=CPdT=dU+PdV.
For 1 mole of an ideal gas, PV=RT⇒PdV=RdT at constant pressure. Substituting:
CPdT=CVdT+RdT⇒CP−CV=R
This is Mayer's relation, showing that CP is always greater than CV by the universal gas constant R.
✓Final answer(A) C_P - C_V = R.
- CBSE 2025Set ANNUAL1 markQ.Answer in one word or one sentence: Write Mayer's equation.
›Reveal solutionSolution
Mayer's relation states Cp - Cv = R for an ideal gas.
For one mole of an ideal gas, the molar specific heat at constant pressure (Cp) exceeds the molar specific heat at constant volume (Cv), because at constant pressure some of the supplied heat must also do work as the gas expands, in addition to raising its internal energy. Using the first law of thermodynamics and the ideal gas equation, it can be shown that this excess exactly equals the universal gas constant R, giving Mayer's equation: Cp - Cv = R.
✓Final answerMayer's equation: Cp - Cv = R.
- CBSE 2024Set ANNUAL1 markMCQQ.The ratio of specific heats (γ) of diatomic gas is:(a) 3/5(b) 5/7(c) 7/9(d) 7/5
›Reveal solutionSolution
Using the equipartition theorem, a diatomic gas has 5 degrees of freedom, giving Cv = 5R/2, Cp = 7R/2, and hence gamma = Cp/Cv = 7/5.
For an ideal gas molecule with f degrees of freedom, the equipartition theorem gives the molar internal energy as U = (f/2)RT, so:
Cv = dU/dT = (f/2)R
Cp = Cv + R = (f/2)R + R = [(f+2)/2]R
gamma = Cp/Cv = (f+2)/f
A diatomic gas molecule (like O2 or N2) has 3 translational degrees of freedom (motion along x, y, z) plus 2 rotational degrees of freedom (rotation about the two axes perpendicular to the bond axis; rotation about the bond axis itself is not counted since the moment of inertia there is negligible). So f = 5.
gamma = (f+2)/f = (5+2)/5 = 7/5 = 1.4
✓Final answerThe correct option is (d) 7/5.
- CBSE 2024Set ANNUAL1 markMCQQ.The total internal energy of a mono atomic gas is(a) a) (1/2)KBT(b) b) (1/3)KBT(c) c) (3/2)KBT(d) d) (5/2)KBT
›Reveal solutionSolution
[!TLDR]
c) (3/2)KBT
Why
A monatomic gas molecule has 3 translational degrees of freedom; by the law of equipartition of energy, average energy per molecule = 3 x (1/2)KBT = (3/2)KBT.
[!ANSWER]
c) (3/2)KBT
- CBSE 2023Set ANNUAL1 markMCQQ.Mayer's formula for the relation between two principal specific heats Cp and Cv of a gas is given by(1) Cv - Cp = R(2) Cp / Cv = R(3) Cp - Cv = R(4) Cv / Cp = R
›Reveal solutionSolution
For one mole of an ideal gas, Mayer's relation states Cp - Cv = R, where R is the universal gas constant.
For an ideal gas, heating it at constant volume only increases its internal energy (all the heat goes into raising temperature), while heating it at constant pressure must also supply energy for the gas to do work as it expands (W = PΔV = RΔT for one mole). This extra work requirement is exactly what makes Cp larger than Cv, and the difference between them equals the gas constant R per mole:
Cp - Cv = R
This is Mayer's formula/relation, derived from the first law of thermodynamics applied to an ideal gas, and used (together with the degrees-of-freedom results) to compute Cp and Cv for monatomic, diatomic, etc. gases.
✓Final answer(3) Cp - Cv = R.
- CBSE 2022Set ANNUAL1 markMCQQ.In the relation PV^γ = constant, γ is:(a) Cp - Cv(b) Cp / Cv(c) Cp . Cv(d) Cp + Cv
›Reveal solutionSolution
The symbol γ in the adiabatic relation PV^γ = constant is the ratio of specific heat at constant pressure to specific heat at constant volume, γ = Cp/Cv.
For a reversible adiabatic process on an ideal gas, combining the first law of thermodynamics (dQ = 0) with the ideal gas equation PV = nRT gives PV^γ = constant, where γ = Cp/Cv. This ratio is always greater than 1 because Cp > Cv (at constant pressure, some of the added heat goes into doing work as the gas expands, so more heat is needed for the same temperature rise). For a monatomic gas γ = 5/3, for a diatomic gas γ = 7/5.
✓Final answerThe correct option is (b) Cp / Cv.
- CBSE 2022Set ANNUAL1 markMCQQ.Match Column A item "Mayer's relation" with the correct item in Column B.(a) kg m^2(b) Poise(c) Cp - Cv = R(d) E = mc^2(e) 1/frequency(f) distance(g) 24 hours
›Reveal solutionSolution
Mayer's relation states Cp - Cv = R (per mole), showing that the extra heat needed at constant pressure over constant volume goes entirely into the work of expansion.
At constant volume, all heat added raises internal energy: dQ = Cv dT. At constant pressure, some heat also does work as the gas expands: dQ = Cp dT = dU + PdV = Cv dT + RdT (using PV = RT for 1 mole). Comparing gives Cp - Cv = R, Mayer's relation.
✓Final answerMayer's relation — (c) Cp - Cv = R.
- CBSE 2022Set ANNUAL1 markMCQQ.(Cp − Cv) is equal to —(a) R x J(b) R − J(c) R/J(d) J/R
›Reveal solutionSolution
Cp − Cv = R/J (molar heat capacities in calorie/heat units).
Mayer's relation for one mole of an ideal gas, when the heat capacities are in energy (joule) units, is Cp−Cv=R.
If instead Cp and Cv are measured in heat units (calories), we divide by the mechanical equivalent of heat J (J ≈ 4.18 J/cal):
Cp−Cv=JR.
✓Final answer(c) R/J.
- CBSE 2020Set ANNUAL1 markMCQQ.The molar specific heat at constant pressure of an ideal gas is (7/2) R. The ratio of specific heat at constant pressure to that at constant volume is:(a) 9/7(b) 7/5(c) 8/7(d) 5/7
›Reveal solutionSolution
Mayer's relation connects Cp and Cv for an ideal gas (Cp - Cv = R); once Cv is found, the ratio gamma = Cp/Cv follows directly.
Given Cp = (7/2) R.
Mayer's relation: Cp - Cv = R, so Cv = Cp - R = (7/2) R - R = (5/2) R.
gamma = Cp / Cv = [(7/2) R] / [(5/2) R] = 7/5.
(This value, gamma = 7/5 = 1.4, corresponds to a diatomic ideal gas, consistent with Cp = 7/2 R.)
✓Final answer(b) 7/5.
- CBSE 2020Set ANNUAL1 markMCQQ.The kinetic energy of one mole of an ideal gas is E = 3/2 RT. The value of Cp is:(a) 0.5 R(b) 0.1 R(c) 1.5 R(d) 2.5 R
›Reveal solutionSolution
The given internal (kinetic) energy expression identifies Cv = (3/2)R directly; adding R (Mayer's relation) gives Cp = (5/2)R = 2.5R.
For one mole of an ideal gas, internal energy (here treated as the total kinetic energy) is E = (3/2) R T, which by definition means:
Cv = dE/dT = (3/2) R.
By Mayer's relation: Cp = Cv + R = (3/2) R + R = (5/2) R = 2.5 R.
(This E = (3/2)RT form is characteristic of a monatomic ideal gas, with 3 translational degrees of freedom.)
✓Final answer(d) 2.5 R.
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