Question

See Periodic Table O See Hint A mixture hot gases expands, changing volume from 245L to 6.82 L, against a constant pressure of 3.22 atm. The P-V work involved is

Answer 1

Here gas expand isobarically(constant pressure process).

Since pressure for this process is a constant, force exerted by the system is constant .

Therefore work done of the system is giv by W = -PΔV

volume changes from 2.45 L to 6.82 L thus, volume change , ΔV = 6.82- 2.45 = 4.37 L

we have Pressur P= 3.22 atm

workdone . W =-( 3.22 * 4.37 )= -14.0714 L.atm

1 liter atmosphere = 101.325 joules

Therefore, 14.07 L atm = (101.325 * 14.07) = 1425.643 J

P-V work invovled is -14.07 L.atm ( or -1425.643 J)

Negative sign indicates workdone by the system.(gas transferred energy to surrounding)

Answer 2

The P-V work is given by P×∆V

= Pressure×(final volume - initial volume)

=3.22×(6.82-2.45)L.atm

=14.07L.atm