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 If half the gas is withdrawn and the temperature is raised to 65.0c, what is the new pressure in the tank in Pa?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

`5.868 * 10^7 Pa`

Step-by-step explanation:

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

`P_1 = 10^8 /2 = 5*10^7 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

`(P_1)/(T_1) = (P_2)/(T_2)`

`P_2 = T_2(P_1)/(T_1) = 338(5*10^7)/(288) = 5.868 * 10^7 Pa`