3.8 Two vessels in thermal contact are connected by a tube...

3.8 Two vessels in thermal contact are connected by a tube with a tap in it. The first vessel has a volume of 2 × 10?³ m³ and contains an ideal gas at a pressure of 2 atm. The second has a volume of 3 × 10?³ m³ and contains the same gas at a pressure of 4 atm. The tap is opened. When the system has reached equilibrium, a) what is the temperature change? b) what is the pressure? c) what is the ratio of the number of moles in the two vessels? [Hint: remember that the total number of moles of gas is constant. Write down the ideal gas equation for each vessel in its initial and final states with the numbers of moles appearing in the equations as unknowns. Then eliminate the n's between the equations.] Adiabatic changes in an ideal gas; difference of principal heat capacities; adiabatic equations

Transcript

00:01 Hello everyone we are going to understand this question here given in the question given volume of first vessel volume of first vessel that is given v1 is equal to volume of first vessel v1 is equal to 2 into 10 to the bar minus 3 meter cube and pressure pressure p1 is equal to 2 a tm volume of second vessel volume of second vessel v2 is equal to 3 into 10 to the power minus 3 3 meter cube and pressure pressure p2 is equal to 4 atm now when valve gets open then total volume of the vesal total volume v3 is equal to v1 plus v2.

02:05 Now we know that as per gas equation, pv is equal to nrt.

02:14 From gas equation, pv is equal to nrt.

02:25 Now, for each vessel initially and finally, for each vessel, we can write p1 v1 is equal to n1 r into t1 pressure of first vessel and volume of first vessel number of moles of first vessel into temperature of first vessel and here r is constant that is gas constant now similarly for vessel 2 for vessel 2 so we can write v2 v2 is equal to n2 into r into t2.

03:18 Now we know that total number of moles of gas will be constant.

03:25 So total number of moles, total number of of moles of gas be constant.

03:49 So we can write n1 plus n2 is equal to n3.

03:54 Again, n1 is p1 v1 upon t1 into r plus p2 v2 upon t2 into r is equal to final volume of gas and final pressure of gas.

04:17 Let it is pressure is p3 and volume is v3 upon final temperature t3.

04:28 Into r let this is equation one but here in the question it is given both vessels are thermally contact vessels are in thermally contact vessels are in thermally connected so initial temperature t1 is equal to t2 is equal to t3 means temperature will remains constant.

05:16 So we can write v1 v1 plus p2 v2 is equal to p3 v3.

05:29 Let this is equation.

05:35 So solution of part a as per condition, as per condition in question, both vessels are in thermally contact.

06:01 Both vessels, both vessels are in thermal.

06:09 Thermal contact or in thermal contact so temperature will remains unchanged so temperature will remain unchanged will remain unchanged hence change in temperature hence change in temperature is zero now solving for part b so here p3 into v3 is equal to p1 v1 plus p2 v2.

07:24 Now substituting the value in this.

07:27 So we can write p3 is equal to p1 v1 plus p2 v2 upon v3 and here v3 is v1 plus v1 so we can write p3 is equal to p1 v1 v2 v2 upon p3 is equal to p1 v1 plus p2 v2 upon b1 plus b2...

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