Answer :
Answer:
CHANGE IN VOLUME -0.013 m3
Explanation:
given data:
we know that pressure remain constant for weighted piston cylinder
P1 =P2
Gas constant R = 259.81 j/Kg- K = 0.2598 kJ/kg K
mass, m1 =m2 = 0.01 kg
P1 = 20kPa
T1 = 100 degree C = 373 K
T2 = 0 degree C= 273 K
FROM IDEAL EQUATION
P1V1 =m RT1
[tex]V1 = \frac{mRT1}{P1}[/tex][tex]V1 = \frac{0.01*0.2598*373}{20}[/tex]
V1 = 0.0484 m3
P2V2 =m RT2
[tex]V2 = \frac{mRT2}{P2}[/tex]
[tex]V2 = \frac{0.01*0.2598*273}{20}[/tex]
V2 = 0.0.3546 m3
CHANGE IN VOLUME OS GIVEN
[tex]\Delta V = V2 -V1[/tex]
= 0.0354 - 0.0484
= -0.013 m3