Answer:
T₂ = 469.73 K = 196.73 °C
Explanation:
First we will find the surface area of rubber stop:
[tex]Area = A= \frac{Volume}{Length} \\\\A = \frac{100\ cm^3}{15\ cm}\\\\A = 6.67\ cm^2 = 6.67 \ x\ 10^{-4}\ m^2[/tex]
Now, we will find the final pressure required to remove the rubber stop:
[tex]Final\ Pressure\ = P_{2} = \frac{Force}{Area}+Atmospheric Pressure \\\\P_{2} = \frac{40\ N}{6.67\ x\ 10^{-4}\ m^2} + 101.3 KPa\\\\ P_{2} = 60000\ Pa + 101.3 KPa = 60\ KPa + 101.3 KPa\\\\P_{2} = 161.3\ KPa[/tex]
Now, we use equation of state:
[tex]\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}[/tex]
for constant volume due to rigid cylinder:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\\T_{2} = \frac{P_{2} T_{1}}{P_{1}}[/tex]
where,
P₁ = initial pressure = 101.3 KPa
P₂ = final pressure = 161.3 KPa
T₁ = Initial Temperature = 22°C = 295 K
T₂ = Final Temperature = ?
Therefore,
[tex]T_{2} = \frac{(161.3\ KPa)(295\ K)}{101.3\ KPa}[/tex]
T₂ = 469.73 K = 196.73 °C
