Arigid, 2.50 l bottle contains 0.458 mol he. the pressure of the gas inside the bottle is 1.83 atm. if 0.713 mol ar is added to the bottle and the pressure increases to 2.05 atm, what is the change in temperature of the gas mixture? the initial temperature of the gas is 122 k. the final temperature of the gas is k.

137 K

The volume is constant, so you can use Gay-Lussac’s Pressure-Temperature Law to calculate the new temperature (you don’t have to use the number of moles).

P1/T1 = P2/T2

Solve for T2: T2= T1 x P2/P1

P1 = 1.83 atm; T1 = 122 K

P2 = 2.05 atm; T2 = ?

∴ T2 = 122 K x (2.05 atm)/(1.83 atm) = 137 K

This result makes sense. Temperature is directly proportional to pressure. You increased the pressure by about 10 %, so the temperature increased by about 10 %.

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.