Question

Consider a gas at (3.420x10^2) kPa in a (7.6100x10^-1) L container at (1.8000x10^1)°C. What will thpressure of the gas be. if it is released into a room with a volume of (1.91x10^4) L and a temperature(1.3000x10^1)°C? Express your answer to three significant digits.

Answer 1

Answer

P2 = 9.84x10^-3 kPa

Explanation

Given:

Pressure 1 = 3.420x10^2 kPa

Volume 1 = 7.6100x10^-1 L

Temperature 1 = 1.8000x10^1 °C

Volume 2 = 1.91x10^4 L

Temperature 2 = 1.3000x10^1°C

Required: To calculate Pressure 2

Solution

We will use the combined gas law to solve this problem

$\frac{P_1V_1}{T_1}\text{ = }\frac{P_2V_2}{T_2}$

Re-arrange to make the subject P2:

P2 = P1V1T2/V2T1

P2 = (3.420x10^2 x 7.6100x10^-1 x 1.3000x10^1)/(1.91x10^4 x 1.8000x10^1)

P2 = 9.84x10^-3 kPa