Question

Temperature of one mole of gas is increased by 1° at constant pressure work done is

Answer 1

Work done = -P(external) * ∆V ( change in volume)
Now, here external pressure is constant.

Only change is the change in volume.

Therefore,
Work done = -P ∆V = -P*(Vf-Vi)

Now, according to gas equation,
PV = nRT
Thus here,
P*V = (1)*R*(T)

Therefore,
P*Vf = R*Tf
and
P*Vi = R*Ti

Hence,
Vf - Vi = R*(Tf-Ti) / P
Here , Tf - Ti = 1°C

Hence,
Work done = -P*(Vf - Vi) = -P * R(1)/P
= - R
Therefore , work done = -R = -8.314 J

Answer 2

Answer:

-R=8.314

Explanation:

work done= -P(external)(Vfinal-Vinitial)

PV=(1)× R×T

PVfinal=RTfinal

PVinitial=RTinitial

W=-P(R(Tfinal-Tinitial)/P

Tfinal-Tinitial=1

W=-R=8.314