One mole of an ideal gas occupies the volume V1, has a temperature T1 and a pressure P1. What is the volume of this gas, taken in an amount of 4 mol, at the same temperature and pressure 2P1.
1. We use the equation of state of an ideal gas (it is also called the Clapeyron-Mendeleev equation):
PV = nRT.
2. We have P1V1 = ν1RT1.
P2V2 = ν2RT2.
We get V2 / V1 = ν2 / 2ν1 = 4/2 * 1 = 2.
V2 = 2 V1.
We have: ν2 = 4 mol. ν1 – 1 mol.
P2 = 2P1.
T1 = T 2.
Answer: the volume of this gas taken in an amount of 4 mol is 2 times larger, at the same temperature and pressure 2P1.
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