Question

A same of gas in a rigid container is at 25.0Cand 1.00atm. What is the pressure of the sample when heated to 220.0C

Answer 1

Answer:

1.654 atm.

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 and V are constant, and have different values of P and T:

(P₁T₂) = (P₂T₁)

• Knowing that:

P₁ = 1.0 atm, T₁ = 25°C + 273 = 298 K,

P₂ = ??? atm, T₂ = 220°C + 273 = 493 K,

• Applying in the above equation

(P₁T₂) = (P₂T₁)

∴ P₂ = (P₁T₂)/(T₁) = (1.0 atm)(493 K)/(298 K) = 1.654 atm.