The ratio γ = (Cp)/(Cv) for a gas mixture consisting of 8 g of helium and 16 g of oxygen is (Physics Olympiad

I. Multiple Choice Questions · Q7

\gamma = \dfrac{C_p}{C_v}

(a)

23/15

(b)

15/23

(c)

27/17

(d)

17/27

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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 $C_p$ .

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 ( $C_v$ ): 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 ( $C_p$ ): Some heat goes into internal energy, but some also goes into the work of expansion against the constant external pressure. So $C_p$ is always larger than $C_v$ for gases (and for most solids/liquids, the difference is tiny because they barely expand).

Note

For an ideal gas, the difference is exactly $C_p - C_v = nR$ , 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:

C_p = \left( \frac{\delta Q}{dT} \right)_p

The subscript $p$ means "at constant pressure." The $\delta Q$ (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:

\delta Q_p = dH

Therefore:

C_p = \left( \frac{\partial H}{\partial T} \right)_p

This is the working definition you'll use in problems: $C_p$ 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 ( $C_{p,m}$ ): heat capacity per mole (units: J mol⁻¹ K⁻¹)
Specific heat capacity at constant pressure ( $c_p$ ): heat capacity per unit mass (units: J kg⁻¹ K⁻¹)

The total heat capacity of a sample is:

C_p = n \cdot C_{p,m} = m \cdot c_p

Why It Matters

In most chemical reactions and physical processes, the system is open to the atmosphere — constant pressure. So $C_p$ is the relevant quantity for:

Calculating enthalpy changes ( $\Delta H = n C_{p,m} \Delta T$ )
Designing calorimeters (like coffee-cup calorimeters that operate at constant pressure) …

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Step 1. Convert masses to moles: helium ( $M=4\ \text{g mol}^{-1}$ ), $\mu_1=8/4=2$ mol; oxygen ( $M=32\ \text{g mol}^{-1}$ ), $\mu_2=16/32=0.5$ mol.

Step 2. Helium is monatomic ( $C_V=\tfrac32R$ , $C_P=\tfrac52R$ ); oxygen is diatomic at normal temperature ( $C_V=\tfrac52R$ , $C_P=\tfrac72R$ ).

Step 3. For the mixture, specific heats add in proportion to moles present: $C_V^{mix}=\mu_1(\tfrac32R)+\mu_2(\tfrac52R)=2(\tfrac32R)+0.5(\tfrac52R)=3R+1.25R=4.25R$ , and $C_P^{mix}=\mu_1(\tfrac52R)+\mu_2(\tfrac72R)=2(\tfrac52R)+0.5(\tfrac72R)=5R+1.75R=6.75R$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The ratio of Molar specific Heats (Cp/Cv) of Monoatomic gas is :$
(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).

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The ratio of Cp and Cv for helium gas is$
(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.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Mayer's formula is (A) C_P - C_V = R (B) C_P/C_V = R (C) C_V - C_P = R (D) C_V/C_P = R$
›Reveal solutionSolution

Mayer's formula states $C_P - C_V = R$ for one mole of an ideal gas.

At constant volume, all the heat supplied goes into raising internal energy: $dQ = C_VdT = dU$ .

At constant pressure, the gas also does work as it expands, so extra heat is needed for the same temperature rise: $dQ = C_PdT = dU + PdV$ .

For 1 mole of an ideal gas, $PV=RT \Rightarrow PdV = RdT$ at constant pressure. Substituting:

…

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CBSE 2025Set ANNUAL1 mark
$Q. Answer in one word or one sentence: Write Mayer's equation.$
›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 e …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The ratio of specific heats (γ) of diatomic gas is:$
(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

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The total internal energy of a mono atomic gas is$
(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 …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. Mayer's formula for the relation between two principal specific heats Cp and Cv of a gas is given by$
(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

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. In the relation PV^γ = constant, γ is:$
(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 int …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Match Column A item "Mayer's relation" with the correct item in Column B.$
(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.

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. (Cp - Cv) is equal to —$
(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 $C_p - C_v = R$ .

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. The molar specific heat at constant pressure of an ideal gas is (7/2) R. The ratio of specific heat at constant pressure to that at constant volume is:$
(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.

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. The kinetic energy of one mole of an ideal gas is E = 3/2 RT. The value of Cp is:$
(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.

…

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Q.The ratio γ=CpCv\gamma = \dfrac{C_p}{C_v}γ=Cv​Cp​​ for a gas mixture consisting of 8 g of helium and 16 g of oxygen is (Physics Olympiad - 2005)

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Q.The ratio $\gamma = \dfrac{C_p}{C_v}$ for a gas mixture consisting of 8 g of helium and 16 g of oxygen is (Physics Olympiad - 2005)

Unlock everything free for 14 days