Answer:0.0704 kg
Step-by-step explanation:
Given
initial Absolute pressure
=210+101.325=311.325
![T_1=25^(\circ)\approx 298 K](https://img.qammunity.org/2020/formulas/physics/high-school/8v0uan0uxym51ky3qwuj511j6a3s74vzyn.png)
![V=0.025 m^3](https://img.qammunity.org/2020/formulas/physics/high-school/g30d27qqwf10qwmc50qu6beqwsxdx5uqun.png)
![T_2=50^(\circ)\approx 323 K](https://img.qammunity.org/2020/formulas/physics/high-school/1a61h3bfkr4znmrfmv7up68dha9quaa0rh.png)
as the volume remains constant therefore
![(P_1)/(T_1)=(P_2)/(T_2)](https://img.qammunity.org/2020/formulas/chemistry/high-school/rk4xg6yrmhego89fatld1qeuepq135sjo6.png)
![(311.325)/(298)=(P_2)/(323)](https://img.qammunity.org/2020/formulas/physics/high-school/2671x5yn6qg3bfn61bq76854foiyutltx3.png)
![P_2=337.44 KPa](https://img.qammunity.org/2020/formulas/physics/high-school/1fr2kmdzjf08785lc16gkscur7mvugxtva.png)
therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass
![m_1=(P_1V)/(RT_1)=(311.325* 0.025)/(0.0287* 298)](https://img.qammunity.org/2020/formulas/physics/high-school/bw2n912zoqtkmv7pqrvq271o01dkzn3pxv.png)
![m_1=0.91 kg](https://img.qammunity.org/2020/formulas/physics/high-school/yyqhc48iv7152p7uiitvwzz453ntd1ao5t.png)
Final mass
![m_2=(P_2V)/(RT_2)=(311.325* 0.025)/(0.0287* 323)](https://img.qammunity.org/2020/formulas/physics/high-school/2juwb6dhf80r434k6fy0x1pf5b3vdthhfm.png)
![m_2=0.839](https://img.qammunity.org/2020/formulas/physics/high-school/r90e2fixiwddfgjsrl3vfl37u5j0905mgs.png)
Therefore
=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back