Question 11: A student forgot to add the reaction mixture to the round bottomed flask at 27° C but instead he/she placed the flask on the flame. After a lapse of time, he realised his mistake, and using a pyrometer he found the temperature of the flask was 477° C. What fraction of air would have been expelled out?

**Answer:** Given, T_{1} = 27 + 273 = 300 K, T_{2} = 477 + 273 = 750 K

We can find the ratio of V_{1} and V_{2} by using the following equation:

`(V_1)/(T_1)=(V_2)/(T_2)`

Or, `(V_1)/(300)=(V_2)/(750)`

Or, `(V_1)/(V_2)=(300)/(750)=2/5`

Out of total volume (after expansion) only 2 parts out of 5 parts are left in flask, i.e. 3 parts are expelled.

Fraction of air expelled `=3/5`

Question 12: Calculate the temperature of 4.0 mol of a gas occupying 5 dm^{3} at 3.32 bar. (R = 0.083 bar dm^{3} K^{-1} mol^{-1}).

**Answer:** Using pV = nRT

Or, `T=(pV)/(nR)`

`=(3.32xx5)/(4xx0.083)=50` K

Question 13: Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

**Answer:** Molar mass of N_{2} = 28 u

So, 28 g of N_{2} has 6.022 × 10^{23} molecules

Since atomic number of nitrogen is 7 hence there are 7 electrons in one atom and 14 electrons in one molecule of N_{2}

So, number of electrons in 28 g N_{2} = 6.022 × 10^{23} × 14

Or, number of electrons in 1.4 g N_{2}

`=(6.022xx10^(23)xx14xx1.4)/(28)=4.2154xx10^(23)`

Question 14: How much time would it take to distribute one Avogadro number of wheat grains, if 10^{10} grains are distributed each second?

**Answer:** Avogadro Number = 6.022 × 10^{23}

Distribution of 10^{10} grains takes 1 second

So, distribution of one Avogadro number of wheat grains will take

`(6.022xx10^(23))/(10^(10))=6.022xx10^(13)` second

`=(6.022xx10^(13))/(60xx60xx24xx365)=1.9xx10^6` year

Question 15: Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm^{3} at 27° C. (R = 0.083 bar dm^{3} K^{-1} mol^{-1}).

**Answer:** Molar mass of O_{2} = 32 g mol^{-1}

So, number of moles in 8 g O_{2} `=8/(32)=0.25` mol

Molar mass of H_{2} = 2 g mol^{-1}

So, number of moles in 4 g H_{2} `=4/2=2` mol

So, total number of moles (n) = 2 + 0.25 = 2.25

Given, V = 1 dm^{3}, T = 27 + 273 = 300 K, R = 0.083 bar dm^{3} K^{-1} mol^{-1}

`p=(nR\T)/V`

`=(2.25xx0.083xx300)/1=56.025` bar

Question 16: Pay load is defined as the difference between the mass of 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 K K^{-1} mol^{-1}).

**Answer:** Let us first calculate the volume of balloon

Volume of sphere `=4/3πr^3`

`=4/3xx(22)/7xx10^3=4190.5` m^{3}

This is the volume of given amount of He filled at 27° C

Now, mass can be calculated as follows:

`pV=nR\T`

Or, `pV=(mR\T)M`

Or, `m=(Mp\V)/(RT)`

`=(4xx1.66xx4190.5)/(0.083xx300)=1117.5` kg

Total mass of balloon = 100 + 1117.5 = 1217.5 kg

Maximum mass of air that can be displaced by balloon to go up = Volume × density

= 4190.5 × 1.2 = 5028.6 kg

So, payload = 5028.6 â€“ 1217.5 = 3811.1 kg

Question 17: Calculate the volume occupied by 8.8 g of CO_{2} at 31.1° C and 1 bar pressure. (R = 0.083 bar K K^{-1} mol^{-1}).

**Answer:** Molar mass of CO_{2} = 44 g

So, number of moles in 8.8 g `=(8.8)/(44)=0.2`

Given, p_{1} = 1 bar, p_{2} = 1 bar, T_{1} = 31.1 + 273 = 304.1 K, T_{2} = 273 K, n_{1} = 0.2, n_{2} = 1

Using pV = nRT

`V=(nR\T)/p`

`=(0.2xx0.083xx304.1)/1=5.04` L

Question 18: 2.9 g of a gas at 95° C occupied the same volume as 0.184 g of dihydrogen at 17° C, at the same pressure. What is the molar mass of the gas?

**Answer:** Given, m_{1} = 2.9 g, M_{1} = ?, T_{1} = 95 + 273 = 368 K, m_{2} = 0.184 g, M_{2} = 2, T_{2} = 17 + 273 = 290 K

Using pV = nRT

`n_1T_1=n_2T_2`

Or, `(m_1)/(M_1)T_1=(m_2)/(M_2)T_2`

Or, `(2.9xx368)/(M_1)=(0.184xx290)/2`

Or, `M_1=(2.9xx368xx2)/(0.184xx290)=40` g mol^{-1}

Question 19: A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

**Answer:** Let us assume that the mixture contains 100 g of gases. Then, weight of H_{2} = 20 g and that of O_{2} = 80 g

Number of moles of H_{2} `=(20)/2=10`

Number of moles of O_{2} `=(80)/(32)=2.5`

Total number of moles = 10 + 2.5 = 12.5

Partial pressure of hydrogen

= No. of mole of H_{2} ÷ No. of total mole × total pressure

`=(10)/(12.5)xx1=0.8` bar

Question 20: What would be the SI unit for the quantity pV^{2}T^{2}/n?

**Answer:** SI unit of pressure = N m^{-2}, volume = m^{3}, temperature = K, n = mol

So, SI unit of pV^{2}T^{2}/n

= (N m^{-2}) × (m^{3})^{2}× K^{2} × mol^{-1}

= N m^{4} K^{2} mol^{-1}

Question 21: In terms of Charlesâ€™ Law, explain why -273° C is the lowest possible temperature.

**Answer:** At -273° C the volume of a gas becomes zero, so it is considered the lowest possible temperature.

Question 22: Critical temperature for carbon dioxide and methane are 31.1° C and -81.9° C respectively. Which of these has stronger intermolecular forces and why?

**Answer:** Liquefaction of carbon dioxide begins at a higher temperature (31.1° C) than liquefaction of methane. This means that molecules of carbon dioxide can more easily coalesce to form liquid compared to the molecules of methane. So, carbon dioxide has stronger intermolecular forces.

Question 23: Explain the physical significance of van der Waals parameters.

**Answer:** The van der Waals equation is as follows:

`(p+(an^2)/(V^2))(V-nb)=nR\T`

Here, a is measure of magnitude of intermolecular forces in a gas and nb is approximately total volume occupied by molecules.

Two assumptions of kinetic theory gases do not hold good and they are as follows:

- There is no force of attraction between molecules of a gas.
- Volume occupied by molecules is negligible compared to the total volume of a gas.

Since this assumptions do not hold good hence gases show deviation from behavior of ideal gases. Had there been no force of attraction between molecules no gas could be liquefied.

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