Answer:0.0704 kg
Explanation:
Given
initial Absolute pressure[tex](P_1)[/tex]=210+101.325=311.325
[tex]T_1=25^{\circ}\approx 298 K[/tex]
[tex]V=0.025 m^3[/tex]
[tex]T_2=50^{\circ}\approx 323 K[/tex]
as the volume remains constant therefore
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
[tex]\frac{311.325}{298}=\frac{P_2}{323}[/tex]
[tex]P_2=337.44 KPa[/tex]
therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass [tex]m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}[/tex]
[tex]m_1=0.91 kg[/tex]
Final mass [tex]m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}[/tex]
[tex]m_2=0.839[/tex]
Therefore [tex]m_1-m_2[/tex]=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back
