The pressure of a gas is 1.0 atm, the volume is 3.0 l, and the temperature is 200 k. a chemist changes one factor while keeping another constant so that the new pressure is 2.0 atm. which of the following could be the new conditions?
a. the final volume is 6.0 l, while temperature is kept constant.
b. the final temperature is 100 k, while volume is kept constant.
c. the final volume is 4.0 l, while temperature is kept constant.
d. the final temperature is 400 k, while volume is kept constant.

The answer to your question is letter: letter D

Explanation:

Data

P1 = 1 atm

V1 = 3 l

T = 200 K

V2 = 2 atm

Formula

          P1V1 = P2V2         (P1/T1) = (P2/T2)             (V1/T1) = (V2/T2)

A. The final volume is 6.0 L, while temperature is kept constant.  This answer is wrong because if the final pressure is 2 atm, then the final volume will be  1.5 l, not 6 l.

B. The final temperature is 100 K, while volume is kept constant.  This answer is incorrect, if the final pressure is 2, then the final temperature is 400°K

C. The final volume is 4.0 L, while temperature is kept constant.  This answer is incorrect, if the pressure increases, then the volumen diminishes.

D. The final temperature is 400 K, while volume is kept constant. This answer is correct, if the final pressure is 2, then the final temperature is 400°K.

D the final temperature is 400 K, while volume is kept constant.

Explanation: Took the quiz :)

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)