Question #145870

A 10.0 L flask contains 0.200 mole of methane CH4, 0.300 mole of hydrogen, 0.400 mole of nitrogen at 25 degree celsius. (a) what is the pressure, in atmosphere inside the flaskn? (b) what is the partial pressure of each component of the mixture of gases.?

Expert's answer

a)the pressure of a mixture of gases is found by Dalton's law:

p=p1+p2+p3=RTV(m1M1+m2M2+m3M3)=8.31×298.1510×10−3(0,2+0.3+0.4)=222986.39=2.22mPa=2.2atmp=p1+p2+p3=\frac{RT}{V}(\frac{m1}{M1}+\frac{m2}{M2}+\frac{m3}{M3})=\frac{8.31\times298.15}{10\times10^{-3}}(0,2+0.3+0.4)=222 986.39=2.22mPa=2.2atm

b)the partial pressure of methane:

p=mM×V×RT=n×RTV=0.2×8.31×298.1510×10−3=49552.53Pa=49.6kPap=\frac{m}{M\times V}\times RT=\frac{n\times RT}{V}=\frac{0.2\times8.31\times298.15}{10\times10^{-3}}=49 552.53Pa=49.6kPa

the partial pressure of hydrogen:

p=mM×V×RT=n×RTV=0.3×8.31×298.1510×10−3=74328.80Pa=74.3kPap=\frac{m}{M\times V}\times RT=\frac{n\times RT}{V}=\frac{0.3\times8.31\times298.15}{10\times10^{-3}}=74328.80Pa=74.3kPa

the partial pressure of nitrogen:

p=mM×V×RT=n×RTV=0.4×8.31×298.1510×10−3=99105.06Pa=99.1kPap=\frac{m}{M\times V}\times RT=\frac{n\times RT}{V}=\frac{0.4\times8.31\times298.15}{10\times10^{-3}}=99105.06Pa=99.1kPa


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