" ReportA. Volume-Pressure Data for GasesPrepare a graph using the following data for a gas sample at 50deg C, as you adjust the size of thecontainer. Pressure (atm) Volume(L) 3.21 8.243.75 7.075.03 5.265.86 4.527.86 3.369.16 2.8912.28 2.1514.30 1.8419.16 1.371. Which is the independent variable? Volume The independent variable is graphed alongthe horizontal (x) axis.2. Which is the dependent variable? Pressure This variable is graphed on the vertical(y) axis.3. You will make two graphs on a single side of graph paper. The data above will begraphed on the upper half. Choose the proper scale for each axis. The axis for volumeshould start at 1 L (or less), and should go up to 8.5 L (or more).The axis for pressureshould range from 3 atm to 20 atm. Choose an appropriate value for the scale of eachaxis so that the data fills most of the upper half of the graph paper.4. Draw and label each axis, and write the value of each variable every 5 or 10 lines.5. Locate each data point, and use a pencil to place a small dot in the appropriate place,circle each data point. Does the data appear to be the linear? Ans. Is NO. It is curved.Use a franch curve or a flexicurveto draw a smooth curve through the data points. Thisline may not go through all of the points. The points may be slightly scattered above andbelow the line or curve.6. Put a title on the graph.The previous data show an inverse relationship between volume and pressure. As thepressure is increased, the volume decreases. Mathematically, this relationship can beexpressed as: Pa1/V;or P = constant (1/V)To determine if the above relationship is correct, and to determine the value of theconstant, a second graph will be prepared.7. Complete the table given by calculating the value of 1/V for each volume measured.Pressure (atm) Volume (L) V-1 (litres -1)3.21 8.24 .1213.75 7.07 0.141 5.03 5.26 0.1905.86 4.52 0.2217.86 3.36 0.2989.16 2.89 0.34612.28 2.15 0.46514.30 1.84 0.54319.16 1.37 0.7308. Construct a graph of inverse volume and pressure on the bottom half of the graphpaper. Put the independent variable on the x-axis, and the independent variable on they-axis. Start at the origin (0,0), and choose the proper scale for each axis. The scaleshould be such that the data fills most of the area on the graph. Label each axis and puta title on the graph.9. Plot the data points by placing a small pencil dot o each data point, circle the datapoints.10. If the data is linear, use the see-through straight edge so that you can see all of the datapoints while you draw the best line through data points. Some points may be off theline.Do not draw several segments to connect the data points. Draw the single best lineto represent the data. If the data is non-linear, use a French curve to make a smoothcurve through the data points. Extend the line pass the data, using a dotted line, untilyou reach one of the axes. Does the line go through the origin ?Yes So, it indicatessimple inverse proportionality between volume and pressure. 11. If you got a straight line for the graph of inverse volume versus pressure, determine theslope of the line to obtain the proportionality constant. The slope is the (change in they)/(change in the x). Choose two points on the line that are on the corners of the gridand spaced fairly far apart. Do not choose data points to determine the slope of thegraph. Indicate the points on the graph used to calculate the slope on your graph. 12.13. Calculation:By graph :A (0.543, 14.30) B (0.298, 7.86)Slope = change in y/change inn x = ( 14.30-7.86)/ (0.543- 0.298) = 6.44/0.245 = 26.286B. Volume-Temperature Data for Gases:Prepare a graph using the following data for a gas at constant pressure, as temperature increases.Temperature(in deg C) Temperature (in K) Volume (mL)11 284 95.325 298 10047 320 107.473 346 116.1159 432 145233 506 169.8258 531 178.11. Which is the independent variable? Temperature.plot this along the horizontal x axis . whichisthe dependent variable? Volume. This should be graphed along the vertical y axis .2. For this graph you will need to determine the intercept. This is the value along the vertical yaxis where the value of x axis is equal to zero. Because you need to determine the intercept,scale the x axis so that it begins at zero. The vertical y axis does not need to start at zero.scale each axis so that the data fills most of the graph paper. Label each axis and write inthe values every 5 or 10 lines. Write the title at the top of the graph.3. Plot the data by making a small dot for each point .circle the data points.4. If the data is linear, use a see through straight edge or ruler to draw the straight line thatbest represent the data. The line may not go through all points .using a dotted line, extendthe line past the data to the intercept, where the line intersects with the vertical y axis.Determine the value of the intercept and clearly indicate the intercept and its value on thegraph.5. Calculate the slope of the line. Choose two points on the line that are on the corners of thegrid, and far apart. Do not use data points to calculate the slope. Note thet the slope willhave units. Show the calculation of the slope below. Be sure to indicate the points to usedcalculate the slope of your graph. 6. Calculation:A ( 176, 520) B ( 94,280)Slope = (Change in y)/(change in x) = (520-280)/(176-94)-1= 240/82 = 2.93 V TThe data should obey the straight line retionship: Y = mx + bWhere x and y are temperature and volume, m is the slope of the line, and b is the intercept. Writethe equation that is represented by your data using V, T, and your slope and intercept. Be sure toinclude any units for the slope and intercept .Equation of a line passing through a point (a,b) and given slope m is ; y – b = m (x-a)So here lets point A (176,520) and slope(m) = 2.93Y – 520 = 2.93 (x – 176)By graphV – 520 = 2.93 (T – 176)T V = 2.93 T + 4.32T This relationship between the volume of a gas and temperature has significance. Use your equationto calculate the lowest theoretical limit of temperature, absolute-zero. This would be thetemperature at which the volume of a gas equals to the zero. A lower temperature would imply anegative volume for the gas, and is meaning-less. Of course a real gas would liquefy or solidify at thisvery low temperature.Use your equation to calculate the temperature at which the volume of the gas equals zero:Now V = 0T 2.93 T + 4.32 = 0T = -0.678 deg C or 272.47KAbsolute zero = 272.47KNow % error = (|experimental value – actual value |/actual value)* 100%% error= (-272.47 – (-273.15))/273.15 *100=0.25%C. Vapour Pressure Curves:There is a table of the vapour pressure of water at various temperatures. You will make two graphs,each on a separate page. The first graph will illustrate the relationship between temperature and thevapour pressure of water. Although the data doesn’t include the boiling point of water, you canextend the graph beyond the data points this is called extrapolation. The line or curve representingthe data is extended beyond the data points using a broken or dotted line.1. The first graph will be of the vapour pressure and temperature data on the table thatfollows. The independent variable goes along the x-axis, and the dependent variable goesalong the y –axis.2. You will extend the graph to beyond the data points to estimate the boiling point of water,so the scale must extend beyond the data on the following table. Choose a scale for thetemperature axis that goes up to 110 deg C.3. Choose a scale for the vapour pressure that goes up to 800 mmHg.4. Label each axis, showing the value every 5 or 10 lines.5. Plot the data points by making small dots for each data point. Circle the data point.6. Put a title on the graph.7. Vapour pressure data is usually curved. Use a French curve or flexicurve to make a smoothline that best represent the data. Extend the line, using a dotted line, past the data points upto a pressure of 800mmHg.8. The normal boiling point of a substance is defined as the temperature at which the vapourpressure equals exactly 1atm or 760 mmHg. Using the extended line on yourgraph,determine the temperature at which the vapour pressure would equal to the 1 atm.Using a straight edge, draw a thin line across the graph at 1 atm (760 mmHg) pressure. Atthe point where extended line and 1 atm line intersect, use a straight edge to draw a fineline to the temperature axis.this temperature is the estimated boiling point of the water.indicate the value for the boiling point of water on the temperature axis of yourgraph.9. Calculation of % error in the estimation of the boiling point ;% error = |boiling point from graph – actual boiling point |/ (actual boiling point)*100%By graph, % error= ((101-100)/100)*100=1%-1 -1 Temp (deg C) Vapour Temp(K) Temp (K ) ln(vapourpressure(mmHg) pressure)0 4.6 273 .00370 1.5265 6.5 278 .00360 1.87110 9.2 283 .00351 2.21915 12.8 288 .00347 2.54920 17.5 293 .00341 2.86225 23.8 298 .00335 3.16930 31.8 303 .00330 3.45935 42.2 308 .00324 3.74240 55.3 313 .00319 4.01245 71.9 318 .00314 4.27550 92.5 323 .00309 4.52755 118.0 328 .00304 4.77060 149.4 333 .00300 5.00665 187.5 338 .00295 5.23470 233.7 343 .00291 5.45475 289.1 348 .00287 5.66780 355.1 353 .00283 5.87285 433.6 358 .00279 6.07290 525.8 363 .00275 6.26595 633.9 368 .00271 6.452Several relationships involving a measured property versus temperature are curved like theprevious graph. Often , if the natural logarithm of the measured property is graphed versusinverse of the temperature (in kelvins), the graph will be linear. In addition, the slope of theline often has physical significance.10. Complete the table above. The third column should be the temperature in Kelvins. Thefourth column should be the inverse of the temperature in K-1, and the last column shouldbe the natural logarithm (ln) of the vapour pressure. Note that when the ln of a quantity istaken, there are no units.11. Use another piece of graph paper to make a graph of ln of vapour pressure versus inversetemperature. Choose a scale for each axis so that the data fills most of the part of the graphpaper. You do not need to start at the origin(0,0),. Label each axis.12. Plot each of the data points, and circle each point. The data should be linear.use a see- through straight edge to draw a straight line that best fits the data. Put a title on the graph.13. Calculate the slope of the line. Choose two points on the line that are n the corners of thegrid and far apart. Do not use the data points to calculate the slope. Indicate the two points used to calculate the slope of the graph. show the calculation below. Note that the ln of Phas no units but 1/T has the units of K-1. As a result, the slope will have units. 14."