The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the differe

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Q. The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the difference between CP and CV for 10 moles of an ideal gas.

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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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The difference $C_P - C_V$ for an ideal gas is $nR$ , independent of the gas and the temperature. For 10 moles, this difference is $10R \approx 83.14\ \text{J K}^{-1}$ .

The relation $H = U + pV$ is the definition of enthalpy. For an ideal gas, $pV = nRT$ , so $H = U + nRT$ . The heat capacities at constant pressure and constant volume are defined as the partial derivatives of enthalpy and internal energy with respect to temperature:

C_P = \left(\frac{\partial H}{\partial T}\right)_p, \quad C_V = \left(\frac{\partial U}{\partial T}\right)_V

The key insight is that for an ideal gas, internal energy $U$ depends only on temperature, not on volume or pressure. This means $\left(\frac{\partial U}{\partial T}\right)_V = \frac{dU}{dT}$ , the same derivative regardless of the constraint. Similarly, enthalpy $H = U + nRT$ also depends only on temperature for an ideal gas, so $\left(\frac{\partial H}{\partial T}\right)_p = \frac{dH}{dT}$ .

Start from the definition of enthalpy: $H = U + pV$ . For an ideal gas, $pV = nRT$ , so $H = U + nRT$ .
Differentiate $H$ with respect to temperature at constant pressure:

C_P = \left(\frac{\partial H}{\partial T}\right)_p = \frac{dU}{dT} + nR

The constant-volume heat capacity is:

C_V = \left(\frac{\partial U}{\partial T}\right)_V = \frac{dU}{dT}

Subtract the two expressions: $C_P - C_V = \left(\frac{dU}{dT} + nR\right) - \frac{dU}{dT} = nR$ …

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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 36 on this concept.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$^\circ$
›Reveal solutionSolution

Total internal energy $U=\left(\tfrac{3}{2}n_1+\tfrac{5}{2}n_2\right)RT=10RT=33.24\,\mathrm{kJ}$ .

For $n_1=5$ moles monoatomic gas, $C_V=\tfrac{3}{2}R$ ; for $n_2=1$ mole rigid diatomic gas, $C_V=\tfrac{5}{2}R$ . Temperature $T=127\,^\circ\mathrm{C}=400\,\mathrm{K}$ . …
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TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
$Q_1$
›Reveal solutionSolution
The ratio $Q_1 : Q_2$ is found by calculating the total heat (latent + sensible) for each case and simplifying. The result is $3 : 8$ .

The problem asks for the ratio of two heats, each involving two stages: melting ice at $0^\circ \mathrm{C}$ into water at $0^\circ \mathrm{C}$ (latent heat), then heating that water to a final temperature (sensible heat). The key is to treat each stage separately and sum them.

Latent heat depends only on mass, not on temperature change. Sensible heat depends on mass, specific heat, and the temperature rise. So for each case, we compute $Q = m L_f + m c \Delta T$ , then take the ratio.

For $Q_1$ (2 g ice to $40^\circ \mathrm{C}$ water):

Mass $m_1 = 2\,\mathrm{g}$ .

Latent heat: $Q_{\text{latent},1} = m_1 L_f = 2 \times 80 = 160\,\mathrm{cal}$ .

Sensible heat (heating water from $0^\circ \mathrm{C}$ to $40^\circ \mathrm{C}$ ): $Q_{\text{sensible},1} = m_1 c \Delta T_1 = 2 \times 1 \times 40 = 80\,\mathrm{cal}$ .

Total: $Q_1 = 160 + 80 = 240\,\mathrm{cal}$ .

For $Q_2$ (4 g ice to $80^\circ \mathrm{C}$ water):

Mass $m_2 = 4\,\mathrm{g}$ .

Latent heat: $Q_{\text{latent},2} = 4 \times 80 = 320\,\mathrm{cal}$ .

Sensible heat (heating from $0^\circ \mathrm{C}$ to $80^\circ \mathrm{C}$ ): $Q_{\text{sensible},2} = 4 \times 1 \times 80 = 320\,\mathrm{cal}$ .

Total: $Q_2 = 320 + 320 = 640\,\mathrm{cal}$ .

Find the ratio: …
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TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
$-10\,^\circ\mathrm{C}$
›Reveal solutionSolution
The key idea is to check whether all the ice melts or not by comparing the heat needed to warm and melt the ice with the heat the water can supply. The final equilibrium mixture has ice and water in the ratio 1:2, so the correct option is (B).

Concept and Intuition

When ice and water at different temperatures are mixed, heat flows from the warmer water to the colder ice. The ice first warms to 0 °C, then may melt if enough heat remains. The water cools to 0 °C. The final state depends on whether the water’s available heat is enough to melt all the ice. We calculate the heat required to bring the ice to 0 °C and then melt it, and compare it with the heat the water can give up by cooling to 0 °C. If the water’s heat is insufficient, only part of the ice melts, and the mixture stays at 0 °C with both ice and water present.

Step-by-step solution

Heat needed to warm the ice from –10 °C to 0 °C Mass of ice $m_i = 80\ \text{g}$ , specific heat of ice $c_i = 0.5\ \text{cal g}^{-1}\text{°C}^{-1}$ .

$Q_1 = m_i c_i \Delta T = 80 \times 0.5 \times (0 - (-10)) = 80 \times 0.5 \times 10 = 400\ \text{cal}.$

Heat needed to melt all the ice at 0 °C Latent heat of fusion $L = 80\ \text{cal g}^{-1}$ .

$Q_2 = m_i L = 80 \times 80 = 6400\ \text{cal}.$
Total heat required to turn all ice into water at 0 °C:
$Q_{\text{ice, total}} = Q_1 + Q_2 = 400 + 6400 = 6800\ \text{cal}.$

Heat the water can supply by cooling from 20 °C to 0 °C Mass of water $m_w = 100\ \text{g}$ , specific heat of water $c_w = 1\ \text{cal g}^{-1}\text{°C}^{-1}$ .

$Q_{\text{water}} = m_w c_w \Delta T = 100 \times 1 \times (20 - 0) = 2000\ \text{cal}.$

Compare the heat available with the heat needed

The water can supply only 2000 cal, but melting all the ice requires 6800 cal. Clearly, not all ice melts. The mixture will end at 0 °C with some ice remaining.

How much ice actually melts?

First, the ice must be warmed to 0 °C, which uses 400 cal. The remaining heat from the water is
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TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
$-20\,^\circ\mathrm{C}$
›Reveal solutionSolution

Sum the four stages — warming ice, melting, warming water, boiling — to get $24528\ \mathrm{J}$ , which is $5840\ \mathrm{cal}$ — option (B).

Concept

Taking $8\ \mathrm{g} = 0.008\ \mathrm{kg}$ from ice at $-20\,^\circ\mathrm{C}$ to steam at $100\,^\circ\mathrm{C}$ needs four heat inputs: raising ice to $0\,^\circ\mathrm{C}$ , melting it, raising water to $100\,^\circ\mathrm{C}$ , and vaporising it.

Solution
$Q_1 = m c_\text{ice}\,\Delta T = 0.008 \times 2100 \times 20 = 336\ \mathrm{J}.$ $Q_2 = m L_f = 0.008 \times 336\times10^3 = 2688\ \mathrm{J}.$
$Q_3 = m c_\text{water}\,\Delta T = 0.008 \times 4200 \times 100 = 3360\ \mathrm{J}.$ …
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TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$1.5$
›Reveal solutionSolution
For an adiabatic process, $TV^{\gamma-1} = \text{constant}$ . With $\gamma = 1.5$ and $T_2 = 2T_1$ , the volume ratio is $V_2/V_1 = 1/2\sqrt{2} \approx 0.3536$ , so the percentage decrease is about $64.6\%$ — but the given options suggest a different interpretation: the problem likely intends a sudden compression as an adiabatic process where $T \propto V^{1-\gamma}$ , leading to a $75\%$ decrease, matching option (D).

Concept & Intuition

The phrase “suddenly compressed” is the key. In thermodynamics, a sudden change means there is no time for heat exchange with the surroundings — the process is adiabatic. For an ideal gas undergoing a reversible adiabatic process, the relation between temperature and volume is
$T V^{\gamma-1} = \text{constant},$
where $\gamma = C_p/C_v$ is the ratio of specific heats. Here $\gamma = 1.5$ , and the temperature doubles. We can directly find how the volume changes.

Step-by-step solution

Write the adiabatic relation For an adiabatic process:

$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}.$
Given $T_2 = 2 T_1$ and $\gamma = 1.5$ , so $\gamma - 1 = 0.5$ .

Substitute and solve for the volume ratio

$T_1 V_1^{0.5} = (2 T_1) V_2^{0.5}.$
Cancel $T_1$ (non-zero):
$V_1^{0.5} = 2 V_2^{0.5}.$
Square both sides:
$V_1 = 4 V_2 \quad \Rightarrow \quad \frac{V_2}{V_1} = \frac{1}{4}.$

Interpret the result …
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TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
$Q. A Carnot engine uses diatomic gas as a working substance. During the adiabatic expansion part of the cycle, if the volume of the gas becomes 32 times its initial volume, then the efficiency of the engine is (A) 100% (B) 75% (C) 50% (D) 25%$
›Reveal solutionSolution
The efficiency of a Carnot engine depends only on the temperatures of the reservoirs. For a diatomic gas undergoing adiabatic expansion with a volume ratio of 32, the temperature ratio is found using the adiabatic relation $TV^{\gamma-1} = \text{constant}$ , with $\gamma = 7/5$ . This gives an efficiency of 75%, so the correct option is (B).

The key idea is that in a Carnot cycle, efficiency is $\eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}$ . The adiabatic expansion step connects the hot and cold temperatures via the volume change. For a diatomic gas, the adiabatic index $\gamma = C_p/C_v = 7/5$ . Using the relation $T V^{\gamma-1} = \text{constant}$ , we can find the temperature ratio from the given volume ratio, and then compute the efficiency.

Recall the Carnot efficiency formula

The efficiency of a Carnot engine is $\eta = 1 - \frac{T_c}{T_h}$ , where $T_h$ is the temperature of the hot reservoir and $T_c$ is the temperature of the cold reservoir. The adiabatic expansion in the cycle connects these two temperatures.

Apply the adiabatic relation for an ideal gas

For a reversible adiabatic process, $T V^{\gamma-1} = \text{constant}$ . Let the initial volume before expansion be $V_1$ and the final volume after expansion be $V_2 = 32 V_1$ . The temperatures at these points are $T_h$ (start of adiabatic expansion) and $T_c$ (end of adiabatic expansion). Thus:

$T_h V_1^{\gamma-1} = T_c V_2^{\gamma-1}$
Rearranging:
$\frac{T_c}{T_h} = \left( \frac{V_1}{V_2} \right)^{\gamma-1}$

Substitute the volume ratio and $\gamma$ for a diatomic gas For a diatomic gas, $\gamma = \frac{7}{5}$ , so $\gamma - 1 = \frac{2}{5}$ . The volume ratio $V_2/V_1 = 32$ , so $V_1/V_2 = 1/32$ . Therefore:
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TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$Q. A Carnot engine uses diatomic gas as a working substance. During the adiabatic expansion part of the cycle, if the volume of the gas becomes 32 times its initial volume, then the efficiency of the engine is (A) 100% (B) 50% (C) 75% (D) 25%$
›Reveal solutionSolution
For a Carnot engine with a diatomic gas, the adiabatic relation $TV^{\gamma-1} = \text{constant}$ links the temperature drop to the volume expansion. Using $\gamma = 7/5$ for a diatomic gas and a 32‑fold volume increase in the adiabatic expansion, the efficiency comes out to 75%.

The key idea is that a Carnot engine’s efficiency depends only on the temperatures of the two reservoirs: $\eta = 1 - T_C/T_H$ . In the adiabatic expansion step, the gas cools from $T_H$ to $T_C$ as its volume increases. For an ideal gas undergoing a reversible adiabatic process, temperature and volume are related by $TV^{\gamma-1} = \text{constant}$ . So the ratio $T_C/T_H$ can be found directly from the volume ratio, without needing any other data.

A diatomic gas at ordinary temperatures has $\gamma = C_p/C_v = 7/5 = 1.4$ . That’s a standard result: $C_v = \frac{5}{2}R$ , $C_p = \frac{7}{2}R$ , so $\gamma-1 = 2/5 = 0.4$ .

Now let’s work through the calculation.

Adiabatic relation for the expansion For a reversible adiabatic process, $T V^{\gamma-1} = \text{constant}$ . If the gas starts at temperature $T_H$ and volume $V_1$ , and ends at $T_C$ and volume $V_2$ , then

$T_H V_1^{\gamma-1} = T_C V_2^{\gamma-1}.$
Rearranging gives
$\frac{T_C}{T_H} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}.$

Plug in the given numbers The volume becomes 32 times its initial value, so $V_2/V_1 = 32$ , hence $V_1/V_2 = 1/32$ . For a diatomic gas, $\gamma-1 = 2/5 = 0.4$ . Therefore

$\frac{T_C}{T_H} = \left(\frac{1}{32}\right)^{0.4}.$

Simplify the exponent Write $32 = 2^5$ . Then …
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TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
$= 3 \times 10^{-2}~\mathrm{Nm}^{-1}$
›Reveal solutionSolution
The energy required to expand a soap bubble is the work done against surface tension to increase its total surface area (considering both inner and outer surfaces). This energy is calculated as the product of surface tension and the change in total surface area. For the given radii, the energy required is $\boxed{528 \times 10^{-6}~\mathrm{J}}$ .

Concept and Intuition

When a liquid surface expands, molecules from the bulk of the liquid move to the surface. To do this, they must overcome the cohesive forces pulling them inwards, which requires energy. This energy is stored as potential energy in the expanded surface, and the work done to create this new surface area is directly proportional to the increase in area. This phenomenon is quantified by surface tension ( $T$ ), which can be defined as the work done per unit increase in surface area.

For a soap bubble, a critical detail is that it has two free surfaces in contact with air: an inner surface and an outer surface. Therefore, when the radius of the bubble increases, both these surfaces expand. The total surface area that changes is twice the geometric surface area of a sphere.

The energy required to increase the surface area of a liquid film by $\Delta A$ is given by:
$W = T \times \Delta A$
Step-by-step Derivation

Identify the given values and convert units:

Surface tension of soap solution, $T = 3 \times 10^{-2}~\mathrm{Nm}^{-1}$ .

Initial radius of the soap bubble, $R_1 = 3~\mathrm{cm} = 3 \times 10^{-2}~\mathrm{m}$ .

Final radius of the soap bubble, $R_2 = 4~\mathrm{cm} = 4 \times 10^{-2}~\mathrm{m}$ .

Calculate the initial total surface area ( $A_1$ ) of the soap bubble:

A spherical soap bubble has two surfaces (inner and outer). The surface area of a single sphere is $4\pi R^2$ . Therefore, the total surface area of a soap bubble is $2 \times 4\pi R^2 = 8\pi R^2$ .

$A_1 = 8\pi R_1^2 = 8\pi (3 \times 10^{-2}~\mathrm{m})^2$

$A_1 = 8\pi (9 \times 10^{-4}~\mathrm{m}^2)$

$A_1 = 72\pi \times 10^{-4}~\mathrm{m}^2$

Calculate the final total surface area ( $A_2$ ) of the soap bubble:

Similarly, for the final radius:

$A_2 = 8\pi R_2^2 = 8\pi (4 \times 10^{-2}~\mathrm{m})^2$

$A_2 = 8\pi (16 \times 10^{-4}~\mathrm{m}^2)$

$A_2 = 128\pi \times 10^{-4}~\mathrm{m}^2$

Calculate the change in total surface area ( $\Delta A$ ):

$\Delta A = A_2 - A_1$ …
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TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$27\,^\circ \mathrm{C}$
›Reveal solutionSolution
The radiation emitted by a black body scales as $T^4$ (Stefan–Boltzmann law). Converting temperatures to Kelvin and taking the ratio gives a factor of $(450/300)^4 = (1.5)^4 = 5.0625$ , so the increase is $406.25\%$ — option (D).

The key idea is Stefan’s law: for a black body, the total power radiated per unit area is $\sigma T^4$ , where $T$ is the absolute temperature in Kelvin. When the temperature changes, the radiated power changes as the fourth power of the ratio of absolute temperatures. A common mistake is to use Celsius directly — that breaks the $T^4$ law because it’s not a ratio scale. Always convert to Kelvin first.

Convert temperatures to Kelvin.

$T_1 = 27^\circ\mathrm{C} + 273 = 300\ \mathrm{K}$

$T_2 = 177^\circ\mathrm{C} + 273 = 450\ \mathrm{K}$

Apply Stefan–Boltzmann law.

The power radiated per second (for the same surface area) is proportional to $T^4$ . So

$\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{450}{300}\right)^4 = (1.5)^4$
Compute: $1.5^2 = 2.25$ , then $2.25^2 = 5.0625$ . Hence $P_2 = 5.0625\,P_1$ .

Find the percentage increase. …
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TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$125^\circ \mathrm{C}$
›Reveal solutionSolution
The key idea is that the power radiated by a black body is proportional to the fourth power of its absolute temperature (Stefan–Boltzmann law). Converting the given Celsius temperature to Kelvin and applying the ratio of powers gives the new emission rate as $512~\mathrm{Wm}^{-2}$ , so the correct option is (C).

Concept and intuition

A black body is an ideal emitter: the power it radiates per unit area depends only on its absolute temperature $T$ (in Kelvin). The Stefan–Boltzmann law states:
$P = \sigma T^4$
where $\sigma$ is the Stefan–Boltzmann constant.

If the temperature changes, the new power is proportional to the fourth power of the new absolute temperature. The problem gives an initial temperature in Celsius and a temperature increase in Kelvin (which is the same as an increase in Celsius degrees). The crucial step is to work entirely in Kelvin.

Step-by-step solution

Convert the initial temperature to Kelvin The initial temperature is $125^\circ\mathrm{C}$ .

$T_1 = 125 + 273 = 398~\mathrm{K}$

Identify the temperature increase The temperature is increased by $398~\mathrm{K}$ . So the new absolute temperature is:

$T_2 = T_1 + 398 = 398 + 398 = 796~\mathrm{K}$

Apply the Stefan–Boltzmann law The initial power per unit area is $P_1 = 32~\mathrm{Wm}^{-2}$ . Since $P \propto T^4$ , we have:

$\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$

Compute the ratio

$\frac{T_2}{T_1} = \frac{796}{398} = 2$
Therefore:

$\frac{P_2}{P_1} = 2^4 = 16$ …
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TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
$^\circ$
›Reveal solutionSolution

Warming the water to $100^\circ\text{C}$ condenses only $40\text{ g}$ of steam, leaving $20\text{ g}$ steam and $400\text{ g}$ water — a ratio of $1:20$ .

Heat needed to raise the water to $100^\circ\text{C}$ .

The $360\text{ g}$ of water at $40^\circ\text{C}$ needs to be heated by $60^\circ\text{C}$ :
$Q_{\text{water}} = m\,c\,\Delta T = 360 \times 1 \times (100-40) = 21600 \text{ cal}.$
Heat available from condensing all the steam.

Condensing the entire $60\text{ g}$ of steam would release
$Q_{\text{steam}} = 60 \times 540 = 32400 \text{ cal}.$
Since $32400 > 21600$ , not all the steam condenses. Equilibrium is reached at $100^\circ\text{C}$ with steam and water coexisting.

Mass of steam that condenses.

Let $m$ grams of steam condense, supplying exactly the heat the water needs: …
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TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
$^\circ$
›Reveal solutionSolution
The rms speed of a gas depends on temperature and molar mass; setting the oxygen rms speed to 75% of nitrogen’s rms speed gives a relation that yields the required temperature. The answer is 87 °C.

The key idea is that the root‑mean‑square speed of gas molecules is given by
$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$
where $T$ is the absolute temperature (in kelvin) and $M$ is the molar mass (in kg mol⁻¹).

We are told that for oxygen, $v_{\text{rms,O}_2}$ is 75% of $v_{\text{rms,N}_2}$ at a known temperature of nitrogen. This lets us set up a ratio that eliminates the gas constant $R$ and directly relates the temperatures and molar masses.

Convert the given nitrogen temperature to kelvin The nitrogen temperature is $287\,^\circ\text{C}$ .

$T_{\text{N}_2} = 287 + 273 = 560\ \text{K}$

Write the rms speed expressions For nitrogen (molar mass $M_{\text{N}_2} = 28\ \text{g mol}^{-1} = 0.028\ \text{kg mol}^{-1}$ ):

$v_{\text{rms,N}_2} = \sqrt{\frac{3R \cdot 560}{0.028}}$
For oxygen (molar mass $M_{\text{O}_2} = 32\ \text{g mol}^{-1} = 0.032\ \text{kg mol}^{-1}$ ) at unknown temperature $T_{\text{O}_2}$ :
$v_{\text{rms,O}_2} = \sqrt{\frac{3R \cdot T_{\text{O}_2}}{0.032}}$

Apply the given condition

$v_{\text{rms,O}_2} = 0.75 \cdot v_{\text{rms,N}_2}$
Substitute the expressions:
$\sqrt{\frac{3R\,T_{\text{O}_2}}{0.032}} = 0.75 \cdot \sqrt{\frac{3R \cdot 560}{0.028}}$

Cancel common factors and square both sides The factor $3R$ cancels from both sides:

$\sqrt{\frac{T_{\text{O}_2}}{0.032}} = 0.75 \cdot \sqrt{\frac{560}{0.028}}$
Squaring:
$\frac{T_{\text{O}_2}}{0.032} = (0.75)^2 \cdot \frac{560}{0.028}$ $\frac{T_{\text{O}_2}}{0.032} = 0.5625 \cdot \frac{560}{0.028}$

Simplify the right‑hand side First compute $\frac{560}{0.028}$ :
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Q.The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the difference between CP and CV for 10 moles of an ideal gas.

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