A 20 L container holds 0.650 mol of He gas at 37 °C at a pressure of 628.3 bar.
A 20 L container holds 0.650 mol of He gas at 37°C at a pressure of 628.3 bar. What will be new pressure inside the container if the volume is reduced to 12 L. The temperature is increased to 177°C and 1.25 mol of additional He gas was added to it?
1 Answers
Given :
V1 = Initial volume = 20 L,
n1 = Initial number of moles = 0.650 mol
P1 = Initial pressure = 628.3 bar
T1 = Initial temperature = 37°C
= 37 + 273.15 K
= 310.15 K
n2 = Final number of moles = 0.650 + 1.25
= 1.90 mol,
V2 = Final volume = 12 L
T2 = Final temperature = 177°C
= 177 + 273.15 K
= 450.15 K,
R = 0.0821 L atm K-1 mol-1
To find :
P = Final pressure
Formula :
PV = nRT
Calculation :
According to ideal gas equation,
P2V2 = n2RT2.
∴ P2 = \(\frac{n_2RT_2}{V_2}\)
= \(\frac{1.90\times 0.0821 \times 450.15}{12}\)
= 5.852 atm.
∴ The final pressure of the gas is 5.852 atm.
[Note : In the above numerical, converting the pressure value to different units, we get : 5.852 atm = 4447.52 torr = 5.928 bar]