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:

Solved Problems

Question 1:

A gas occupies one litre under atmospheric pressure. What will be the volume of the same amount of gas under 750 mm of Hg at the same temperature?

Solution.

Given V1 = 1 litre

P1 = 1 atm

V2 = ?

P2 = 750 / 760 atm

Using

P1V1 = P2V2

We get

1 × 1 = 750 / 760 × V2

V2 = 1.0133 litre = 1013.3 ml