krit.club logo

States of Matter: Gases and Liquids - Liquefaction of Gases

Grade 11ICSEChemistry

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

πŸ”‘Concepts

β€’

Liquefaction of gases is the process by which a gas is converted into a liquid state, typically achieved by lowering the temperature and increasing the pressure.

β€’

Critical Temperature (TcT_c): The temperature above which a gas cannot be liquefied, no matter how much pressure is applied. For CO2CO_2, the TcT_c is 31.1∘C31.1^\circ C (304.25K304.25 K).

β€’

Critical Pressure (PcP_c): The minimum pressure required to liquefy a gas at its critical temperature (TcT_c).

β€’

Critical Volume (VcV_c): The volume occupied by one mole of a gas at its critical temperature (TcT_c) and critical pressure (PcP_c).

β€’

Andrews Isotherms: Thomas Andrews studied the Pβˆ’VP-V relationship of CO2CO_2 at various temperatures. He found that below TcT_c, the isotherms show a horizontal portion where gas and liquid coexist in equilibrium.

β€’

Continuity of State: There is no sharp boundary between the gaseous and liquid states when a substance is transitioned along a path that avoids the liquid-vapor equilibrium region (passing above TcT_c and PcP_c).

β€’

Van der Waals Constants and Critical Constants: The critical constants TcT_c, PcP_c, and VcV_c can be expressed in terms of the van der Waals constants 'aa' (intermolecular forces) and 'bb' (molecular size).

β€’

Joule-Thomson Effect: When a gas under high pressure is allowed to expand into a region of low pressure through a porous plug or a fine orifice under adiabatic conditions, its temperature decreases (except for H2H_2 and HeHe at room temperature).

πŸ“Formulae

Tc=8a27RbT_c = \frac{8a}{27Rb}

Pc=a27b2P_c = \frac{a}{27b^2}

Vc=3bV_c = 3b

PcVcRTc=38=0.375\frac{P_c V_c}{R T_c} = \frac{3}{8} = 0.375

(P+aV2)(Vβˆ’b)=RT\left( P + \frac{a}{V^2} \right) (V - b) = RT

πŸ’‘Examples

Problem 1:

Calculate the critical temperature (TcT_c) of a gas if the van der Waals constants are a=3.60 dm6 atm molβˆ’2a = 3.60 \, \text{dm}^6 \, \text{atm} \, \text{mol}^{-2} and b=0.040 dm3 molβˆ’1b = 0.040 \, \text{dm}^3 \, \text{mol}^{-1}. (Take R=0.0821 L atm Kβˆ’1 molβˆ’1R = 0.0821 \, \text{L} \, \text{atm} \, \text{K}^{-1} \, \text{mol}^{-1})

Solution:

Using the formula Tc=8a27RbT_c = \frac{8a}{27Rb}: Tc=8Γ—3.6027Γ—0.0821Γ—0.040T_c = \frac{8 \times 3.60}{27 \times 0.0821 \times 0.040} Tc=28.80.088668T_c = \frac{28.8}{0.088668} Tcβ‰ˆ324.8 KT_c \approx 324.8 \, \text{K}

Explanation:

The critical temperature is determined by the balance of intermolecular attractions (represented by aa) and the finite volume of molecules (represented by bb).

Problem 2:

Determine the critical volume (VcV_c) for a gas if its van der Waals constant bb is 0.05 L molβˆ’10.05 \, \text{L} \, \text{mol}^{-1}.

Solution:

Using the relation Vc=3bV_c = 3b: Vc=3Γ—0.05 L molβˆ’1V_c = 3 \times 0.05 \, \text{L} \, \text{mol}^{-1} Vc=0.15 L molβˆ’1V_c = 0.15 \, \text{L} \, \text{mol}^{-1}

Explanation:

Critical volume depends only on the effective size of the gas molecules, which is represented by the constant bb.

Liquefaction of Gases - Revision Notes & Key Formulas | ICSE Class 11 Chemistry