States of Matter
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5.21. In terms of Charles’ law explain why -273°C is the lowest possible temperature.
5.21. In terms of Charles’ law explain why -273°C is the lowest possible temperature.
5.21. At -273°C, volume of the gas becomes equal to zero, i.e., the gas ceases to exist.
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5.20. What would be the SI unit for the quantity PV2T2/n?
5.20. What would be the SI unit for the quantity PV2T2/n?
5.20. SI unit of pressure, P = Nm-2
SI unit of volume = m3
Si unit of temperature, T = K
SI unit of number of moles, n = mol
Thus, SI unit of pV2T2/n = (Nm-2) (m3)2 (K) 2mol = Nm4K2mol-1
Upvote 5.19. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
5.19. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
5.19. As the mixture H2 and O2 contains 20% by weight of dihydrogen, therefore,
If H2 = 20g, then O2 = 80g
No. of moles of H2 = 20/2 = 10 moles
No. of moles of O2 = 80/32 = 2.5 moles
Partial pressure of H2 = [No. of moles of H2/ (No. of moles of H2 + No. of moles of O2)] x
Ptota
Upvote 5.18. 2.9 g of a gas at 95°C occupied the same volume as 0.184 g of hydrogen at 17°C at the same pressure. What is the molar mass of the gas?
5.18. 2.9 g of a gas at 95°C occupied the same volume as 0.184 g of hydrogen at 17°C at the same pressure. What is the molar mass of the gas?
5.18. Given, P1 = P2 and V1 = V2
We know that P1V1 = P2V2
Or, n1RT1 = n2RT2
i.e., n1T1 = n2T2
Substituting n =
Upvote 5.17. Calculate the volume occupied by 8.8 g of CO2 at 31.1 °C and 1 bar pressure. R = 0.083 bar LK-1 mol-1
5.17. Calculate the volume occupied by 8.8 g of CO2 at 31.1 °C and 1 bar pressure. R = 0.083 bar LK-1 mol-1
5.17. No. of moles of CO2 = Given mass of CO2 / Molar mass
= 8.8g / 44g mol-1 = 0.2 mol
Pressure of CO2 = 1 bar
RR = 0.083 bar dm3 K–1 mol–1
T = 2
Upvote 5.16. Pay load is defined as the difference between the mass of the displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C (Density of air = 1.2 kg m-3 and R = 0.083 bar dm3 K-1 mol-1).
5.16. Pay load is defined as the difference between the mass of the displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C (Density of air = 1.2 kg m-3 and R = 0.083 bar dm3 K-1 mol-1).
5.16. Radius of the balloon = 10 m
Therefore, volume of the balloon = (4/3)? r3 = (4/3) x (22/7) x (10 m)3
= 4190.5 m3
Volume of He filled at 1.66 bar and 27 °C = 4290.5 m3
To calculate the mass of He,
PV = nRT = (w/M) RT, where M is molar mass of He i.e. 4 g per mole or 4 x 10-3 kg mol-1
=> w =&nbs
Upvote 5.15. Calculate the total pressure in a mixture of 8g of oxygen and 4g of hydrogen confined in a vessel of l dm3 at 27°C. R = 0.083 bar dm3 K-1 mol-1.
5.15. Calculate the total pressure in a mixture of 8g of oxygen and 4g of hydrogen confined in a vessel of l dm3 at 27°C. R = 0.083 bar dm3 K-1 mol-1.
5.15. Molar mass of O2 = 32 g/mol
It means, 8 g of O2 has 8/32 mol = 0.25 mol
Molar mass of H2 = 2 g/mol
It means, 4 g of H2 has 4/2 mol = 2 mol
Therefore, total number of moles, n = 2 + 0.25 = 2.25 mol
Given, V = 1dm3, T = 27°C = 300 K, R = 0.083 bar dm3 K-1 mol-1
Applying PV = nRT,
P = nRT / V
&nb
Upvote 5.14. How much time would it take to distribute one Avogadro number of wheat grains if 1010 grains are distributed each second?
5.14. How much time would it take to distribute one Avogadro number of wheat grains if 1010 grains are distributed each second?
5.14. Time taken to distribute 1010 wheat grains = 1s
Time taken to distribute Avogadro number of wheat grains = (1s x 6.022 x 1023) / 1010
= 6.022 x 1013 s
= (6.022 x 1013 / 60 x 60 x 24 x 365) year
= 1.9 x 106 years
Upvote 5.13. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
5.13. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
5.13. Molecular mass of N2 = 28g
28 g of N2 has No. of molecules = 6.022 x 1023
1.4 g of N2 has No. of molecules = (6.022 x 1023 x 1.4 g)/28 g= 3.011 x 1022 molecules.
Atomic No. of Nitrogen (N) = 7
1 molecule of N2 has electrons = 7 x 2 = 14
3.011
Upvote 5.12. Calculate the temperature of 4.0 moles of a gas occupying 5 dm3 at 3.32 bar (R = 0.083 bar dm3 K-1 mol-1).
5.12. Calculate the temperature of 4.0 moles of a gas occupying 5 dm3 at 3.32 bar (R = 0.083 bar dm3 K-1 mol-1).
5.12. Given,
P= 3.32 bar
V= 5 dm3
n= 4 mol
R= 0.083 bar dm3 K-1 mol-1
PV = nRT
Or T = PV / nR = 3.32 x 5 dm3 / (4.0 mol x 0.083 bar dm3 K-1 mol-1)
= 50 K
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