Physics Q97 – AP-EAMCET 2025

Home Physics ThermodynamicsQuestion #503

When a polyatomic gas is heated at constant pressure, the percentage of heat given to the gas that is converted into external work is (Ratio of the specific heat capacities of the gas =

\frac{4}{3}

)

Clue 1 of 3: Thermodynamics of Gases

Understand the definitions of heat supplied at constant pressure ( $Q_{p}$ ) and work done ( $W$ ) by a gas in an isobaric process.

🧠 Full Solution Path

Step 1: Identify Given Information and Goal✦ Active

The problem states that a polyatomic gas is heated at constant pressure (isobaric process). We are given the ratio of specific heat capacities, $γ = \frac{C_{p}}{C_{v}} = \frac{4}{3}$ . We need to find the percentage of heat given to the gas that is converted into external work, which is $\frac{W}{Q_{p}} \times 100 %$ , where $W$ is the work done and $Q_{p}$ is the heat supplied at constant pressure.

💡 Teacher's Secret Hint

Remember that for an isobaric process, the pressure remains constant, which simplifies the calculation of work done.

Step 2: Express Work Done and Heat Supplied in Terms of Specific Heats○ Expand

For an isobaric process, the heat supplied to $n$ moles of gas for a temperature change $Δ T$ is given by:

Q_{p} = n C_{p} Δ T

The work done by the gas in an isobaric process is $W = P Δ V$ . Using the ideal gas law $P V = n R T$ , we have $P Δ V = n R Δ T$ . So, the work done is:

W = n R Δ T

💡 Teacher's Secret Hint

Ensure you use molar specific heats ( $C_{p}$ , $C_{v}$ ) and the universal gas constant ( $R$ ) when dealing with $n$ moles of gas.

Step 3: Calculate the Ratio of Work Done to Heat Supplied○ Expand

Now, we can find the ratio $\frac{W}{Q_{p}}$ :

\frac{W}{Q_{p}} = \frac{n R Δ T}{n C_{p} Δ T} = \frac{R}{C_{p}}

We know Mayer's relation: $C_{p} - C_{v} = R$ . Substituting $R$ into the ratio:

\frac{W}{Q_{p}} = \frac{C_{p} - C_{v}}{C_{p}} = 1 - \frac{C_{v}}{C_{p}}

We are given $γ = \frac{C_{p}}{C_{v}}$ . Therefore, $\frac{C_{v}}{C_{p}} = \frac{1}{γ}$ . Substituting this into the expression:

\frac{W}{Q_{p}} = 1 - \frac{1}{γ}

💡 Teacher's Secret Hint

This derived formula, $\frac{W}{Q_{p}} = 1 - \frac{1}{γ}$ , is a useful shortcut for isobaric processes.

Step 4: Substitute the Given Value of

γ

and Calculate the Percentage○ Expand

Given $γ = \frac{4}{3}$ . Substitute this value into the ratio:

\frac{W}{Q_{p}} = 1 - \frac{1}{4 / 3} = 1 - \frac{3}{4} = \frac{1}{4}

To express this as a percentage, multiply by $100 %$ :

Percentage = \frac{1}{4} \times 100 % = 25 %

💡 Teacher's Secret Hint

Always double-check the units and ensure the final answer is in the requested format (percentage in this case).

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 97 – AP-EAMCET 2025

Clue 1 of 3: Thermodynamics of Gases

🧠 Full Solution Path