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 dа ta:

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