1.calculate the kelvin temperatures of the water and record your answers in the data table. 2.find the change in the volume of air in the flask from your data and record in data table. 3.use the equation v1 / t1 = v2 / t2 to calculate the expected volume of air when cooled in tap water. 4.how do the expected final volume and the actual final volume compare? 5.what is the significance of elevating or lowering the flask until the water level in the flask is even with the water level in the beaker or container? 6.construct a graph of the data. plot the volume of the gas at room temperature in tap water and in ice water on the y axis. plot the kelvin temperature on the x-axis. print out graph paper for your plot. (click here for graph paper.) 7.extend the plotted line downward until it crosses the temperature axis. this process of extending a graph beyond the experimental data is called extrapolation. 8.at which temperature is the line predicted to cross the x-axis? 9.at which temperature did the line actually cross the x-axis? 10.account for any deviation betweeen the predicted temperature line extrapolation and the actual line extrapolation. 11.real world chemistry - explain why bottled gas containers are equipped with a relief valve?

When measuring volume of air in the flask at the first temperature, a volume of 250 mL was recorded, known as V1. The temperature of the air in the flask in boiling water was recorded as 99ᵒC, known as T2. In order to find the correct calculations, 99ᵒC has to be converted to Kelvin by adding 273. The first temperature in Kelvin is 372K. The value of V1/T1, can be found by putting 250/372. This comes to a total of 0.67. The volume of the air in the flask of the second temperature was 177 mL, known as V2. The temperature of the air in the cooled flask is 7ᵒC, known as T2. 7ᵒC has to be converted to Kelvin by adding 273 which comes to a final total of 280K. The value of V2/T2, found by putting 177/280 comes to a total of 0.63. The near equality in numbers can be attributed to Charles Law. Charles Law states that “as temperature increases, so does the volume of a gas sample when the pressure is held constant”. The result of V1/T1 and V2/T2 were very close to each other. This is due to the fact that this experiment was done in a closed system. In Charles Law, if there is a closed system the two ratios should have equal numbers. This is why it can be expected for the ratio numbers to be very equal.

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.