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Thermodynamics - Isothermal and Adiabatic Processes

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An Isothermal process is a thermodynamic process in which the temperature of the system remains constant (T=constantT = \text{constant}, ΔT=0\Delta T = 0).

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For an ideal gas undergoing an isothermal process, the internal energy remains constant (ΔU=0\Delta U = 0). According to the First Law of Thermodynamics, Q=WQ = W.

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Isothermal processes are typically very slow and require the system to be in a perfectly conducting container placed in a heat reservoir.

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An Adiabatic process is one in which no heat enters or leaves the system (Q=0Q = 0).

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Adiabatic processes usually occur very suddenly or rapidly, or within a perfectly insulated container, such that there is no time for heat exchange.

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In an adiabatic expansion, the system does work at the expense of its internal energy, leading to a decrease in temperature (W=−ΔUW = -\Delta U).

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The ratio of specific heats is denoted by γ=CpCv\gamma = \frac{C_p}{C_v}. For monoatomic gases γ≈1.67\gamma \approx 1.67, and for diatomic gases γ≈1.40\gamma \approx 1.40.

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The slope of an adiabatic P−VP-V curve is γ\gamma times steeper than the slope of an isothermal P−VP-V curve at the same point: (dPdV)adia=γ(dPdV)iso\left(\frac{dP}{dV}\right)_{\text{adia}} = \gamma \left(\frac{dP}{dV}\right)_{\text{iso}}.

📐Formulae

PV=constant (Isothermal process)PV = \text{constant} \text{ (Isothermal process)}

Wiso=nRTln⁡(V2V1)=2.303nRTlog⁡10(V2V1)W_{\text{iso}} = nRT \ln\left(\frac{V_2}{V_1}\right) = 2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)

PVγ=constant (Adiabatic process)PV^\gamma = \text{constant} \text{ (Adiabatic process)}

TVγ−1=constantTV^{\gamma-1} = \text{constant}

P1−γTγ=constantP^{1-\gamma}T^\gamma = \text{constant}

Wadia=P1V1−P2V2γ−1=nR(T1−T2)γ−1W_{\text{adia}} = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}

ΔU=nCvΔT\Delta U = nC_v\Delta T

💡Examples

Problem 1:

A certain volume of dry air at STPSTP is expanded adiabatically to three times its original volume. Find the final pressure. (Given γ=1.4\gamma = 1.4 for air).

Solution:

Initial pressure P1=1 atmP_1 = 1 \text{ atm}, Initial volume =V1= V_1, Final volume V2=3V1V_2 = 3V_1. For an adiabatic process, P1V1γ=P2V2γP_1V_1^\gamma = P_2V_2^\gamma. Therefore, P2=P1(V1V2)γ=1×(13)1.4≈0.215 atmP_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = 1 \times \left(\frac{1}{3}\right)^{1.4} \approx 0.215 \text{ atm}.

Explanation:

In an adiabatic expansion, the pressure drops more significantly than in an isothermal expansion because both the volume increase and the temperature decrease contribute to the pressure drop.

Problem 2:

Two moles of an ideal gas at 300K300 K are compressed isothermally from a volume of 4L4 L to 2L2 L. Calculate the work done on the gas. (Use R=8.314J/mol⋅KR = 8.314 J/mol\cdot K).

Solution:

Using the formula W=nRTln⁡(V2V1)W = nRT \ln\left(\frac{V_2}{V_1}\right), we have n=2n = 2, R=8.314R = 8.314, T=300T = 300, V1=4V_1 = 4, V2=2V_2 = 2. W=2×8.314×300×ln⁡(0.5)≈4988.4×(−0.693)≈−3457JW = 2 \times 8.314 \times 300 \times \ln(0.5) \approx 4988.4 \times (-0.693) \approx -3457 J.

Explanation:

The negative sign indicates that work is done on the gas during compression. Since the process is isothermal, this energy is released as heat to the surroundings to maintain a constant temperature.

Isothermal and Adiabatic Processes - Revision Notes & Key Formulas | ICSE Class 11 Physics