Answer:
CHANGE IN VOLUME -0.013 m3
Step-by-step 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
![V1 = (mRT1)/(P1)](https://img.qammunity.org/2020/formulas/engineering/college/j267t9rdf1k8yrcxvduvfnxtoymkaltazi.png)
![V1 = (0.01*0.2598*373)/(20)](https://img.qammunity.org/2020/formulas/engineering/college/tgcfzibkipsnuk0k0wuyso2vek7ythn27g.png)
V1 = 0.0484 m3
P2V2 =m RT2
![V2 = (mRT2)/(P2)](https://img.qammunity.org/2020/formulas/engineering/college/6qm7dxqcyurx07ngbthaccob9dgnykybl6.png)
![V2 = (0.01*0.2598*273)/(20)](https://img.qammunity.org/2020/formulas/engineering/college/9r8q9rbvpc8dshi8xp31a1f8fkpzdd09qm.png)
V2 = 0.0.3546 m3
CHANGE IN VOLUME OS GIVEN
![\Delta V = V2 -V1](https://img.qammunity.org/2020/formulas/engineering/college/l11ahfyl39t9jta5qsxomz5jg23958z5c7.png)
= 0.0354 - 0.0484
= -0.013 m3