1.00 mole of an ideal gas at STP is cooled to -41°C while the
pressure is increased to 805 mmHg. What is the new volume
of the gas in liters?

V₂ = 18.13 L

Explanation:

Given data:

Mole of gas = 1 mol

Initial temperature = 273 K

Initial pressure = 1 atm

Final volume = ?

Final temperature = -41°C (-41+273 = 232 K)

Final pressure = 805 mmHg (805/760 = 1.05 atm)

Solution:

First of all we will calculate the initial volume of gas.

PV = nRT

V = nRT/P

V = 1 mol × 0.0821 mol.L/atm.K × 273 K / 1 atm

V = 22.4 L/atm / 1 atm

V = 22.4 L     ( initial volume)

Now we will determine the final volume by using equation,

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Now we will put the values.

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1 atm × 22.4 L ×  232 K / 273 K × 1.05 atm

V₂ = 5196.8 atm .L. K / 286.65 atm.K

V₂ = 18.13 L

The liquid droplets appearing on the cupper side of test tube are the water molecules that evaparak on heating and condense on touching the cooler surface of upper part of the test tube