A glass sphere is filled to full volume with a gas. The pressure of the gas inside the sphere is 30.0 atm, and the temperature is 30.0°C. The sphere is taken outside on a cold day. The temperature of the gas decreases to 10.0°C. What is the new pressure of the gas? Assume that the volume is constant.

28atm

Explanation:

Using Gay lussac's law equation as follows:

P1/T1 = P2/T2

Where;

P1 = initial pressure (atm)

T1 = initial temperature (K)

P2 = final pressure (atm)

T2 = final temperature (K)

Based on the information provided in this question;

P1 = 30.0 atm, T1 = 30.0°C, P2 = ?, T2 = 10.0°C

NOTE: Absolute temperature i.e. Kelvin is required for this law

T1 = 30°C + 273K = 303K

T2 = 10°C + 273K = 283K

Using P1/T1 = P2/T2

30/303 = P2/283

Cross multiply

P2 × 303 = 30 × 283

303P2 = 8490

P2 = 8490/303

P2 = 28.02

New pressure of the gas = 28atm

i had this on my unit activity, i can you if you are a plato student. are you a plato student?

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"