Two liters of `N_(2)` at `0^(@)C` and `5` atm pressure is expanded isothermally against a constant external pressure of `1` atm untill the pressure of gas reaches `1` atm. Assuming gas to be ideal, claculate the work of expansion.

Two liters of `N_(2)` at `0^(@)C` and `5` atm pressure is expanded iso

1.	Two liters of `N_(2)` at `0^(@)C` and `5` atm pressure is expanded isothermally against a constant external pressure of `1` atm untill the pressure of gas reaches `1` atm. Assuming gas to be ideal, claculate the work of expansion.
Answer» Given `P_(1)=5` atm, `V_(1)=2` litre, `P_(2)=1` atm, `V_(2)=?` We know that - if T is constant then `P_(1)V_(1)=P_(2)V_(2)` `rArr V_(2)=(P_(1)V_(1))/(P_(2))=(5xx2)/(1)=10` litre `W=-P_(ext)Delta V=-1(V_(2)-V_(1))=-1(10-2)=-8` litre atm `=-8xx101.3=-810.4` joule

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Two liters of `N_(2)` at `0^(@)C` and `5` atm pressure is expanded isothermally against a constant external pressure of `1` atm untill the pressure of gas reaches `1` atm. Assuming gas to be ideal, claculate the work of expansion.

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