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

d

explanation: