Home
Class 11
CHEMISTRY
Boyle temperature of vaiour gases are gi...

Boyle temperature of vaiour gases are given below :
`{:("Gas",T_(B)(K),),(H_(2),117,),(He,23,),(CH_(4),498,),(O_(2),406,):}`
Which gas can be liquefied most easily ?

A

`H_(2)`

B

`CH_(4)`

C

`He`

D

`O_(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine which gas can be liquefied most easily from the given data, we will analyze the boiling temperatures of the gases provided. ### Step-by-Step Solution: 1. **Understanding Boyle Temperature**: - The Boyle temperature (Tb) is the temperature at which a gas behaves ideally at a given pressure. It is also related to the critical temperature of the gas, which indicates how easily a gas can be liquefied. 2. **Given Data**: - The boiling temperatures of the gases are as follows: - Hydrogen (H₂): 117 K - Helium (He): 23 K - Methane (CH₄): 498 K - Oxygen (O₂): 406 K 3. **Identifying the Highest Boiling Temperature**: - From the given data, we can see the boiling temperatures: - H₂: 117 K - He: 23 K - CH₄: 498 K - O₂: 406 K - The highest boiling temperature is for Methane (CH₄) at 498 K. 4. **Conclusion**: - Since the gas with the highest boiling temperature is more likely to have a higher critical temperature, it can be liquefied more easily. Therefore, Methane (CH₄) can be liquefied most easily among the given gases. ### Final Answer: - The gas that can be liquefied most easily is **Methane (CH₄)**.
To determine which gas can be liquefied most easily from the given data, we will analyze the boiling temperatures of the gases provided. ### Step-by-Step Solution: 1. **Understanding Boyle Temperature**: - The Boyle temperature (Tb) is the temperature at which a gas behaves ideally at a given pressure. It is also related to the critical temperature of the gas, which indicates how easily a gas can be liquefied. 2. **Given Data**: ...
Promotional Banner

Similar Questions

Explore conceptually related problems

Boyle's temperature of various gases are given below: Which can be liquefied most easily

Gases posses characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases. {:("Gases",H_(2),He,O_(2),N_(2)),("Critical temperature in kelvin",33.2,5.3,154.3,126):} From the above data what would be the order of liquefaction of these gases ? Start writing the order from the gas liquefying first

Critical temperatures for A, B, C and D gases are 25^(@)C, 10^(@)C, -80^(@)C and 15^(@)C respectively. Which gas will be liquefied more easily ?

For real gases, van der Waals' equation is written as (P+(an^(2))/(V^(2))) (V-nb)= nRT where a and b are van der Waals' constants. Two sets of gases are: (I) O_(2), CO_(2), H_(2) and He(II) CH_(4), O_(2) and O_(2) and H_(2) The gases given in set I in increasing order of b and gases given in set II in decreasing order of a are arranged below. Select the correct order from the following:

For real gases, van der Waals' equation is written as (P+(an^(2))/(V^(2))) (V-nb)= nRT where a and b are van der Waals' constants. Two sets of gases are: (I) O_(2), CO_(2), H_(2) and He(II) CH_(4), O_(2) and O_(2) and H_(2) The gases given in set I in increasing order of b and gases given in set II in decreasing order of a are arranged below. Select the correct order from the following:

The van der Waals' constant 'a' for the gases O_(2), N_(2), NH_(3) "and " CH_(4) are 1.36, 1.39, 4.17 and 2.253 respectively, the gas which can be most easily liquefied is :

The value of van der Waals constant a for the gases O_(2) , N_(2) , NH_(3) , and CH_(4) are 1.360 , 1.390 , 4.170 , and 2.253 L^(2) atm mol^(-2) , respectively. The gas which can most easily be liquefied is

The value of van der Waals constant a for the gases O_(2) , N_(2) , NH_(3) , and CH_(4) are 1.360 , 1.390 , 4.170 , and 2.253 L^(2) atm mol^(-2) , respectively. The gas which can most easily be liquefied is

On the basis fo data given below predict which of the following gases shows least adsorption on a definite amount of charcoal? {:("Gas" ,,CO_(2),,SO_(2),,CH_(4),,H_(2)),("Critical temp./K",,304,,630,,190,,33):}

The essential conditions for liquefaction of gases were discovered by Andrews in 1869 as a result of his study of pressure-volume-temperature relationship for CO_(2) . It was found that above a certain temperature, it was impossible to liquefy a gas whatever the pressure was applied. The temperature below which the gas can be liquefied by the application of pressure alone is called critical temperature (Tc). The pressure required to liquefy a gas at this temperature is called the critical pressure (Pc). The volume occupied by one mole of the substance at the critical temperature and pressure is called critical volume. Critical constants are related with van der waals' constant as follows: V_( c) = 3b, P_( c) =a/(27b^(2)), T_( c) =(8a)/(27 Rb) The values of critical volumes of four gases A, B, C and D are 0.025L, 0.312L, 0.245L and 0.432L respectively. The gas with larger molecular diameter will be :