2.00 l of nitrogen gas at a temperature of 210 k are cooled down until the new volume is 1.25 l. if the pressure and number of gas particles do not change, to what temperature was the gas cooled?
131 k
1.19 x 10-2 k
7.62 x 10-3 k
336 k
answer

131 K

Explanation:

The only variables are volume and temperature, so we can use Charles' Law:

V1/T1 = V2/T2

Data:

V1 = 2.00 L; T1  = 210 K

V2 =  1.25 L; T2 = ?

Calculation:

2.00/210 = 1.25/T2

Multiply each side by the lowest common denominator (210T2)

2.00T2 = 1.25 ×210

2.00T2 = 262.5

       T2 = 262.5/2.00 = 131 K

The gas was cooled to 131 K.

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.