Question

A vessel of 25 L contains 20 g of ideal gas X at 300 K. The pressure eas 20 g of ideal gas X at 300 K. The pressure exerted by the gas is 1 atm. 20 g of idealgas Y is added to the vessel keeping the same temperature. Total preser keeping the same temperature. Total pressure became 3 atm. Upon furtheraddition of 20 g ideal gas Z the presurre became 7 atm. Answer the following questionsequation is applicable on mixture of ideal gases)[Take R = 1/12 L.atm/mol K]5. Find the molar mass of gas X:(A) 20 g(B) 10 g(C) 30 g(D) 5g& Identify the correct statement(s):1.gas Y is lighter than gas XII. gas Z is lighter than gas Y(A) I only(B)ll only(C) Both I and II(D) None of the statementsFind the average molar mass of the mixture of gases X,Y and Z:(A) 40/7(B) 50/7(C) 20(D) 60/7​

Answer 1

Answer:

The ideal gas equation :

PV=nRT  

     

      =  

M

m

​  

RT

Given values for ideal gas X,

V=25L

m=20g

T=300K

P=1atm

Substitute values in the above equation :

1×25=  

M

20

​  

×  

12

1

​  

×300.

M=  

25×12

20×300

​  

 

Thus, the molar mass is M=20g/mol.

Explanation: