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प्रश्न
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres
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उत्तर १
The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:
PV = nRT
Where,
R is the universal gas constant = 8.314 J mol–1 K–1
n = Number of moles = 1
T = Standard temperature = 273 K
P = Standard pressure = 1 atm = 1.013 × 105 Nm–2
`:.V= (nRT)/T`
`= (1xx8.314xx273)/(1.013xx10^5)`
`= 0.0224 m^3`
= 22.4 litres
Hence, the molar volume of a gas at STP is 22.4 litres.
उत्तर २
For one mole of an ideal gas, we have
`PV = RT => V = "RT"/P`
Putting `R = 8.31 J mol^(-1) K^(-1)`, T = 273 K and `P = 1 " atmosphere" = 1.013 xx 10^5 Nm^(-2)`
`:. V = (8.31 xx 273)/(1.013xx 10^5) = 0.0224 m^3`
`= 0.0224 xx 10^6 cm^3 = 22400 ml` [`1 cm^3 = 1ml`]
संबंधित प्रश्न
The figure shows the plot of PV/T versus Pfor 1.00×10–3 kg of oxygen gas at two different temperatures.
(a) What does the dotted plot signify?
(b) Which is true: T1 > T2 or T1 < T2?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00 ×10–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.)
An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32 u)
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