A can of butane with a pressure of 2 atm at 283k has its pressure increased to a new pressure of 5 atm. What is the temperature of the new container?

New temperature T2 = 707.5 K (Approx.)

Explanation:

Given:

Old pressure P1 = 2 atm

Old temperature T1 = 283 K

New Pressure P2 = 5 atm

Find:

New temperature T2

Computation:

Using Gay-Lussac law;

P1 / T1 = P2 / T2

So,

2 / 283 = 5 / T2

New temperature T2 = 707.5 K (Approx.)

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

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