Question

A 22.4-l sample of nitrogen at 3.65 atm and 22°c is simultaneously expanded to 57.4 l and heated to 38°c. what is the new pressure of the gas?

Answer 1

1.5 atm.

Step-by-step explanation:

• We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

• If n is constant, and have different values of P, V and T:

(P₁V₁T₂) = (P₂V₂T₁)

• Knowing that:

P₁ = 3.65 atm, V₁ = 22.4 L, T₁ = 22°C + 273 = 295 K,

P₂ = ??? atm, V₂ = 57.4 L, T₂ = 38°C + 273 = 311 K,

• Applying in the above equation

(P₁V₁T₂) = (P₂V₂T₁)

∴ P₂ = (P₁V₁T₂)/(V₂T₁) = (3.65 atm)(22.4 L)(311 K)/(57.4 atm)(295 L) = 1.5 atm.