A1.54l gas bulb in a chemistry laboratory contains oxygen gas at 21 degrees c and 758 torr. the air conditioning in the laboratory breaks down ( 's morehouse) and the temperature rises to 31 degrees. calculate the final pressure of the gas given this increase in temperature

784 torr

Explanation:

Using  Pressure Law; at constant volume,

P∝ T at constant volume.

P= kT ; meaning

P1/T1 = P2/T2

P1= 758 torr

P2= ?

T1= 21 ° C or (273+21)K

   = 294°K

T2= 31 ° C or (273+31)K

    =  304°K

Substituting into equation (1),

We have:

758/294 = P2/304

P2 = 758 x 304/294

    =   783.78torr

    ≈ 784 torr

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j