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प्रश्न
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cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J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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उत्तर १
Volume of oxygen, V1 = 30 litres = 30 × 10–3 m3
Gauge pressure, P1 = 15 atm = 15 × 1.013 × 105 Pa
Temperature, T1 = 27°C = 300 K
Universal gas constant, R = 8.314 J mole–1 K–1
Let the initial number of moles of oxygen gas in the cylinder be n1.
The gas equation is given as:
`P_1V_1 = eta_1RT_1`
`:.n_1 = P_1V_1/RT_1`
`= (15.195 xx10^5 xx 30 xx 10^(-3))/((8.314) xx 300) = 18.276`
Where,
m1 = Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
∴m1 = n1M = 18.276 × 32 = 584.84 g
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume, V2 = 30 litres = 30 × 10–3 m3
Gauge pressure, P2 = 11 atm = 11 × 1.013 × 105 Pa
Temperature, T2 = 17°C = 290 K
Let n2 be the number of moles of oxygen left in the cylinder.
The gas equation is given as:
P2V2 = n2RT2
`:. n_2= (P_2V_2)/(RT_2)`
`= (11.143 xx 10^5 xx 30 xx 10^(-3))/8.314 xx 290 = 13.86`
But `n_2 = m_2/M`
Where,
m2 is the mass of oxygen remaining in the cylinder
∴m2 = n2M = 13.86 × 32 = 453.1 g
The mass of oxygen taken out of the cylinder is given by the relation:
Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder
= m1 – m2
= 584.84 g – 453.1 g
= 131.74 g
= 0.131 kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.
उत्तर २
Initial Volume, `V_1 = 30 litre = 30 xx 10^3 cm^3`
`= 30 xx 10^3 xx 10^(-6) m^3 = 30 xx 10^(-3) m^3`
Initla Pressure, P_1 = 15 atm
`= 15 xx1.013 xx 10^5 N m^(-2)`
Inital temperature, `T_1 = (27 + 273) K = 300 K`
Initial number of moles
`mu_1 = (P_1V_1)/RT_1 = (15xx1.013 xx10^5 xx 30xx10^(-3))/(8.31 xx 300) = 18.3`
Final Pressure, `P_2 = 11 atm`
`= 11 xx 1.013 xx 10^5 N m^(-2)`
Final Volume, V`_2 = 30 litre = 30 xx 10^(-3) cm^3`
Final temperature, `T_2 = 17 + 273 = 290 K`
Final number of moless, `mu_2 = (P_2V_2)/(RT_2) = (11xx 1.013xx 10^5xx 30xx10^(-3))/(8.31 xx 290) = 13.9`
Number of moles takesout of cyclinder = 18.3 - 13.9 = 4.4
Mass of gas taken out of cylinder = 4.4 xx 32 g = 140.8 g = 0.141 kg
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.)
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic).
Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Oxygen is filled in a closed metal jar of volume 1.0 × 10−3 m3 at a pressure of 1.5 × 105Pa and temperature 400 K. The jar has a small leak in it. The atmospheric pressure is 1.0 × 105 Pa and the atmospheric temperature is 300 K. Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalise with the surrounding.
Match the following:
|
|
Column A |
Column B |
|
(a) |
cm3 |
(i) Pressure |
|
(b) |
Kelvin |
(ii) Temperature |
|
(c) |
Torr |
(iii) Volume |
|
(d) |
Boyle's law |
(iv) `"V"/"T" = ("V"_1)/("T"_1)` |
|
(a) |
Charles's law |
(v) `"PV"/"T" = ("P"_1 "V"_1)/"T"_1` |
|
|
|
(vi) PV = P1V1 |
Fill in the blanks:
If the temperature is reduced to half, ………….. would also reduce to half.
A gas occupies 500 cm3 at a normal temperature. At what temperature will the volume of the gas be reduced by 20% of its original volume, the pressure is constant?
Name or state the following:
The standard pressure of a gas in cm. of mercury corresponding to one atmospheric pressure.
The average energy per molecule is proportional to ______
Estimate the average thermal energy of a helium atom at room temperature (27 °C).
Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197g mole–1.
The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is ______.
At room temperature, a diatomic gas is found to have an r.m.s. speed of 1930 ms-1. The gas is ______.
For a wave, y = 0.0002 sin`[2pi(110"t"-x/3)+pi/3]` is travelling in a medium. The energy per unit volume being transferred by wave if density of medium is 1.5 kg/m3, is ______.
