Question

1. A cylinder of ideal gas is closed by an 8 kg movable piston of area 60cm. The atmospheric pressure is
100 kPa. When the gas is heated from 30° Cto 100° C the piston rises 20 cm. The piston is then fastened
in the place and the gas is cooled back to 30° C IfAQ, is the heat added to the gas during heating and
AQ2 is the heat lost during cooling, find the difference.​

Answer 1

Answer : B

Solution :

b. The gas pressure

=weight of piston Area of cross-section+atm. pressure

=8×9.860×10−4+1.00×105N/m2

=1.13×105N/m2

During the heat pressure, the internal energy is changed by ΔU1 and work ΔW1 is done.

Therefore, ΔQ1=ΔU1+ΔW1=ΔU1+PdV

Δ1+(1.13×105)(0.02×60×10−4)

ΔU1+136J

During the cooling process, no work is done as volume is constant, ΔW=0

Hence, ΔQ2=ΔU2, But ΔU2 is negative as the temperature decreases, and since the gas returns to its original temperature, ΔU2=−ΔU1

Hence

[ΔQ1−|ΔQ2|]=(ΔU1+136−ΔU1)=136J