A 25 mol sample of an ideal gas with a molar specific heat...

A 2.5 mol sample of an ideal gas with a molar specific heat Cv = 5/2R always starts at pressure 1.50 * 10^5 Pa and temperature 350 K. The gas is compressed adiabatically to 200 kPa. a) Determine the final pressure (Pf, in kPa). b) Determine the final volume (Vf, in L). c) Determine the final temperature (Tf, in K). d) Determine the change in internal energy of the gas (Î”Eint, in J). e) Determine the energy added to the gas by heat (Q, in J). f) Determine the work done on the gas (W, in J).

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00:01 Here cv for this gas is 5 by 2 r.

00:08 This implies that cp equals to 7 by 2 r and cp by cv equals to gamma equals to cp by cv will be one point.

00:21 Now initial volume is given as 2 .5 into 22 .4 liter is equals to 56 into 10 to the power minus 3 meter cube.

00:37 Now initial pressure is given as 1 .5 into 10 to the power 5 pascal and temperature given as 350 kelvin.

00:50 Final pressure is also given as 200 kilo pascal.

00:56 In the first part we are asked what is the final pressure.

01:00 Final pressure is already given.

01:02 So pf equals to 200 kilo pascal.

01:08 Now part b we are asked what is the final volume.

01:14 For adiabatic process we know p1 v1 to the power gamma equals to p2 v2 to the power gamma.

01:21 So from this we have p2 equals to p1 by p2 to the power 1 by gamma to v1.

01:30 This is equals to 1 .5 divided by 2 into 1 over 1 .4 into v1 is 56 into 10 to the power minus 3 the unit of meter cube.

01:49 This is v2.

01:50 From this we will have v2 equals to final volume 0 .045 meter cube.

02:00 Now in part c we are asked what is the final temperature.

02:05 Now for adiabatic process we know that final temperature we can calculate as p2 equals to p1 by p2 to the power 1 minus gamma by gamma into t1...

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