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Thermodynamics - Thermodynamic State Variables and Equation of State

Grade 11CBSEPhysics

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

🔑Concepts

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A Thermodynamic System is a collection of a large number of molecules whose state is defined by macroscopic variables like Pressure (PP), Volume (VV), and Temperature (TT).

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State Variables are physical quantities that describe the equilibrium state of a system. They are classified into Intensive and Extensive variables.

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Intensive Variables are independent of the size or mass of the system. Examples include Pressure (PP), Temperature (TT), and Density (ρ\rho).

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Extensive Variables depend on the size or mass of the system. Examples include Volume (VV), Mass (MM), and Internal Energy (UU).

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An Equation of State is a mathematical relationship between the state variables. For an ideal gas, the equation is PV=nRTPV = nRT.

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Thermodynamic Equilibrium occurs when the macroscopic variables (P,V,TP, V, T) of a system are uniform throughout and do not change with time. This implies Mechanical, Thermal, and Chemical equilibrium.

📐Formulae

PV=nRTPV = nRT

n=mM=NNAn = \frac{m}{M} = \frac{N}{N_A}

PV=NkBTPV = N k_B T

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

ρ=PMRT\rho = \frac{PM}{RT}

💡Examples

Problem 1:

A gas occupies a volume of 2.0 m32.0 \text{ m}^3 at a pressure of 1.5×105 Pa1.5 \times 10^5 \text{ Pa} and a temperature of 300 K300 \text{ K}. If the volume is reduced to 1.0 m31.0 \text{ m}^3 and the temperature is increased to 400 K400 \text{ K}, calculate the final pressure P2P_2.

Solution:

Using the combined gas law derived from the equation of state: P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Substitute the given values: P1=1.5×105 Pa,V1=2.0 m3,T1=300 KP_1 = 1.5 \times 10^5 \text{ Pa}, V_1 = 2.0 \text{ m}^3, T_1 = 300 \text{ K} V2=1.0 m3,T2=400 KV_2 = 1.0 \text{ m}^3, T_2 = 400 \text{ K}

Rearrange for P2P_2: P2=P1V1T2V2T1P_2 = \frac{P_1 V_1 T_2}{V_2 T_1} P2=(1.5×105)×2.0×4001.0×300P_2 = \frac{(1.5 \times 10^5) \times 2.0 \times 400}{1.0 \times 300} P2=12×107300P_2 = \frac{12 \times 10^7}{300} P2=4.0×105 PaP_2 = 4.0 \times 10^5 \text{ Pa}

Explanation:

The problem uses the equation of state for a fixed amount of gas (constant nn). By relating the initial and final states, we can find the unknown pressure variable.

Problem 2:

Calculate the difference in volume between a gas at 450 cm3450 \text{ cm}^3 and a compressed state of 125 cm3125 \text{ cm}^3.

Solution:

To find the change in the extensive variable (Volume): 450−125325\begin{array}{r} 450 \\ - 125 \\ \hline 325 \end{array}

The change in volume is 325 cm3325 \text{ cm}^3.

Explanation:

Volume is an extensive state variable. The difference is calculated using simple subtraction of the magnitudes.