Question

Gas is confined in a tank at a pressure of 1.0 x 10^8 Pa and a temperature of 15.0c. If half the gas is withdrawn and the temperature is raised to 65.0c, what is the new pressure in the tank in Pa?

Answer 1

Answer:

$5.868 \times 10^7 Pa$

Explanation:

If half the gas is drawn then pressure would have dropped by half

$P_1 = 10^8 /2 = 5\times10^7 Pa$

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

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

$P_2 = T_2\frac{P_1}{T_1} = 338\frac{5\times10^7}{288} = 5.868 \times 10^7 Pa$