Answer:
[tex]5.868 \times 10^7 Pa[/tex]
Explanation:
If half the gas is drawn then pressure would have dropped by half
[tex]P_1 = 10^8 /2 = 5\times10^7 Pa[/tex]
Assuming ideal gas, if temperature rises from 15c (T1 = 15 + 273 = 288 K) to 65 c (T2 = 65 + 273 = 338 K), then we have the following equation for ideal gas
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]
[tex]P_2 = T_2\frac{P_1}{T_1} = 338\frac{5\times10^7}{288} = 5.868 \times 10^7 Pa[/tex]
