An ideal diatomic gas contracts in an isobaric process from...

An ideal diatomic gas contracts in an isobaric process from 1.15 m3 to 0.560 m3 at a constant pressure of 1.30 ✕ 105 Pa. If the initial temperature is 455 K, find the work done on the gas, the change in internal energy, the energy transfer Q, and the final temperature. HINT (a) the work done on the gas (in J) J (b) the change in internal energy (in J) J (c) the energy transfer Q (in J) J (d) the final temperature (in K) K

Transcript

00:01 Hello students, here given that an ideal diatomic molecule undergoes an isobaric process.

00:07 So isobaric process means it occurs in a constant pressure and here the volume at initial stage is 1 .05 meter cube and this contracts to a final volume of 0 .540 meter cube and this happens at constant pressure of 1 .20 into 10 power 5 pascal and the initial temperature t1 here is 455 kelvin.

00:36 So for this case we have to find the work done on the gas w and the change in internal energy delta u, energy transfer through the process q and the final temperature t2.

00:50 So first let's find the work done on the gas.

00:55 Work done w for an isobaric process is given by pressure multiplied by the change in volume that is v2 minus v1.

01:09 V2 minus v1 and now if we substitute the values we can get the value of work done.

01:18 Here the pressure is 1 .20 into 10 power 5, v2 is 0 .540 minus v1 is 1 .05 and we get the work done that is equal to minus 6 .12 into 10 power 4 joules.

01:40 Here negative sign indicates that the work done by the gas and for the second part the change in internal energy for an isobaric process is given by delta u is equal to 5 by 2 nrt.

02:02 Since it is a diatomic molecule we are using this value and here n is the number of moles and this is calculated by using ideal gas equation.

02:13 So by ideal gas equation that is pv is equal to nrt.

02:20 Here we use initial conditions to find the n that is pv1 by rt1...

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