Question #77756
80 cm3 of hydrogen are collected at 15oC and 75 cm of Mercury pressure. What is it's volume at s.t.p?
Expert's answer
First, we will define STP as T_0=273.15°K, p_0=〖10〗^5 Pa. Then, we can make equations for both states of gas, using the ideal gas law: pV=RT, where p – is pressure, V – is volume, - is amount of substance, R – is gas constant and T – is temperature.
pV=RT
p_0 V_0=RT_0
pV/(p_0 V_0 )=RT/(RT_0 )
pV/(p_0 V_0 )=T/T_0
V_0=(pVT_0)/(p_0 T)
p=75 cm m.p.=〖10〗^4 Pa,V=80 cm^3=8×〖10〗^(-5) m^3,T=15℃=288.15°K
V_0=(〖10〗^4×8×〖10〗^(-5)×273.15)/(〖10〗^5×288.15)=218.52/(〖10〗^6×28.815)≈7.58×〖10〗^(-6) m^3=7.58 cm^3
pV=RT
p_0 V_0=RT_0
pV/(p_0 V_0 )=RT/(RT_0 )
pV/(p_0 V_0 )=T/T_0
V_0=(pVT_0)/(p_0 T)
p=75 cm m.p.=〖10〗^4 Pa,V=80 cm^3=8×〖10〗^(-5) m^3,T=15℃=288.15°K
V_0=(〖10〗^4×8×〖10〗^(-5)×273.15)/(〖10〗^5×288.15)=218.52/(〖10〗^6×28.815)≈7.58×〖10〗^(-6) m^3=7.58 cm^3
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