The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. A cylinder contains oxygen at a temperature of 320 degrees Kelvin and a pressure of

25 atmospheres in a volume of 120 liters. Find the pressure if the volume is decreased to 110

liters and the temperature is increased to 335 degrees Kelvin.

Hint: Look at the equation V = k* Then use the initial values to solve for k. Then plug in the

values you know for the second set of values and solve for your unknown.

Pressure=22.55 atmospheres

Step-by-step explanation:

Let

V=volume of the ideal gas

P=pressure

T=temperature

Volume varies directly with temperature and inversely with pressure

V=kT/P

t=320°K

P=25 atmospheres

V=120 liters

V=kT/P

120=k*320/25

120=320k/25

120×25=320k

3000=320k

k=300/320

=9.375

k=9.375

V=110 liters

p=?

t=335°K

V=kT/P

110=9.375*335/p

110=3,140.625/p

110p=3,140.625

P=3,140.625/110

=28.55

P=22.55 atmospheres

351.23

step-by-step explanation:

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p