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Thermal Properties of Matter - Ideal-gas Equation and Absolute Temperature

Grade 11CBSEPhysics

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

🔑Concepts

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An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions. Real gases behave like ideal gases at high temperatures and low pressures.

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The Ideal Gas Law relates the pressure (PP), volume (VV), and absolute temperature (TT) of a gas using the equation PV=nRTPV = nRT, where nn is the number of moles and RR is the universal gas constant.

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Boyle's Law states that for a fixed amount of gas at constant temperature, pressure is inversely proportional to volume: P∝1VP \propto \frac{1}{V}.

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Charles' Law states that for a fixed amount of gas at constant pressure, volume is directly proportional to the absolute temperature: V∝TV \propto T.

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Gay-Lussac's Law states that for a fixed amount of gas at constant volume, pressure is directly proportional to the absolute temperature: P∝TP \propto T.

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Absolute Zero (0 K0 \text{ K}) is the temperature at which the volume and pressure of an ideal gas would theoretically become zero. It is equal to −273.15∘C-273.15^{\circ}\text{C}.

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The Kelvin scale is the absolute temperature scale. The relationship between Celsius (tt) and Kelvin (TT) is given by T=t+273.15T = t + 273.15.

📐Formulae

PV=nRTPV = nRT

PV=NkBTPV = N k_B T

kB=RNA≈1.38×10−23 J K−1k_B = \frac{R}{N_A} \approx 1.38 \times 10^{-23} \text{ J K}^{-1}

R≈8.314 J mol−1 K−1R \approx 8.314 \text{ J mol}^{-1} \text{ K}^{-1}

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

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

💡Examples

Problem 1:

Calculate the volume occupied by 1 mole1 \text{ mole} of an ideal gas at STP (Standard Temperature and Pressure, where P=1.013×105 PaP = 1.013 \times 10^5 \text{ Pa} and T=273.15 KT = 273.15 \text{ K}).

Solution:

Using the Ideal Gas Equation PV=nRTPV = nRT, we solve for VV: V=nRTPV = \frac{nRT}{P} Substituting the values: V=1×8.314×273.151.013×105V = \frac{1 \times 8.314 \times 273.15}{1.013 \times 10^5} V≈0.0224 m3=22.4 LV \approx 0.0224 \text{ m}^3 = 22.4 \text{ L}

Explanation:

At STP, one mole of any ideal gas occupies a molar volume of approximately 22.422.4 liters.

Problem 2:

A gas at 27∘C27^{\circ}\text{C} and 1 atm1 \text{ atm} pressure is contained in a vessel. If the temperature is raised to 127∘C127^{\circ}\text{C} while the volume remains constant, what will be the new pressure?

Solution:

First, convert temperatures to Kelvin: T1=27+273=300 KT_1 = 27 + 273 = 300 \text{ K} T2=127+273=400 KT_2 = 127 + 273 = 400 \text{ K} Since volume is constant, we use Gay-Lussac's Law: P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2} P2=P1×T2T1=1×400300P_2 = P_1 \times \frac{T_2}{T_1} = 1 \times \frac{400}{300} P2=1.33 atmP_2 = 1.33 \text{ atm}

Explanation:

According to Gay-Lussac's law, pressure is directly proportional to absolute temperature when volume is constant. An increase in temperature leads to an increase in pressure.

Problem 3:

Find the difference in temperature between 500 K500 \text{ K} and the standard room temperature of 25∘C25^{\circ}\text{C} in Celsius.

Solution:

First, convert 500 K500 \text{ K} to Celsius: t=500−273=227∘Ct = 500 - 273 = 227^{\circ}\text{C} Now, perform the subtraction: 227−25202\begin{array}{r} 227 \\ - 25 \\ \hline 202 \end{array} The difference is 202∘C202^{\circ}\text{C}.

Explanation:

Temperature differences are calculated by ensuring both values are in the same unit. 500 K500 \text{ K} is equivalent to 227∘C227^{\circ}\text{C}.