Answer:
a) V1 = 5 ft^3
V2 = 2 ft^3
b) n = 1.378
Step-by-step explanation:
Given data:
mass of gas = 4 lb
starting point
![p_1 = 15 lbf/in^2](https://img.qammunity.org/2021/formulas/physics/high-school/45hmtugfp87qv80uqcreoajhezt1uzr2pr.png)
![v_1 = 1.25 ft^3/lb](https://img.qammunity.org/2021/formulas/physics/high-school/wh8607h5j0ww2u4t4wa6pivod9ixvunfc3.png)
end point
![p_2 = 60 lbf/in^2](https://img.qammunity.org/2021/formulas/physics/high-school/7zpwra5glezw54cgt7dov2ylp2xf9r1sp4.png)
![v_2 = 0.5 ft^3/lb](https://img.qammunity.org/2021/formulas/physics/high-school/1glf7ubtlx3r8khtmp8bwfvx1d5qfms6n3.png)
Assume gas to be ideal
a) volume at point 1
![= v_1 * m](https://img.qammunity.org/2021/formulas/physics/high-school/zxr14ryomlkcpimtlmh75rp5jhgapvzcyz.png)
![V_1 = 1.25 * 4 = 5 ft^3](https://img.qammunity.org/2021/formulas/physics/high-school/pspfpsxpqj4pljqqogc7913jb5yzkaclt2.png)
volume at point 2
![= v_2 * m](https://img.qammunity.org/2021/formulas/physics/high-school/lg4tleaihxvu0bf15ecbslh3692b3ijvx2.png)
![V_2 = 0.5 * 4 = 2 ft^3](https://img.qammunity.org/2021/formulas/physics/high-school/w8nc7cvou4soeon4d68ntp8d29x70flwk0.png)
b) from ideal gas equation we have following equation
![(P_1)/(P_2) = [(V_2)/(V_1)]^n](https://img.qammunity.org/2021/formulas/physics/high-school/lcnnjgyzuz5iygbicyha98b7qv855028em.png)
taking log on both side of equation
![ln [(P_1)/(P_2)] = n * ln [(V_2)/(V_1)]](https://img.qammunity.org/2021/formulas/physics/high-school/brvtsoy5xop0lck31pgirr7ndcttu5gj3m.png)
solving for n
![n = ( ln((15)/(53)))/( ln((2)/(5)))](https://img.qammunity.org/2021/formulas/physics/high-school/ibznkr2f0odftq8ge40wz0lsnu53mr6vbh.png)
![n = (-1.262)/(-0.916)](https://img.qammunity.org/2021/formulas/physics/high-school/7u5bh0oj0ge29rqst1rbl705lv996obx0j.png)
n = 1.378