Question

A rigid metal tank containing a sample of argon kept at 4.5atm is moved from room temperature (298K) to an oven that raises its temperature to 377K.

Answer 1

5.7 atm

Step-by-step explanation:

The question is missing, but I think it must be about the new pressure of the argon gas.

Step 1: Given data

• Initial pressure (P₁): 4.5 atm

• Initial temperature (T₁): 298 K

• Final pressure (P₂): ?

• Final temperature (T₂): 377 K

Step 2: Calculate the final pressure of the argon gas

There is a direct relationship between pressure and temperature. Assuming ideal behavior and constant volume (rigid tank), we can calculate the final pressure of the gas using Gay-Lussac's law.

P₁/T₁ = P₂/T₂

P₂ = P₁ × T₂/T₁

P₂ = 4.5 atm × 377 K/298 K = 5.7 atm