At the sea level, the dry air mass percentage composition is given as nitrogen gas: 70.0 , oxygen gas: 27.0 and argon gas: 3.0 . If total pressure is 1.15 atm , then calculate the ratio of following respectively:
(i) partial pressure of nitrogen gas to partial pressure of oxygen gas
(ii) partial pressure of oxygen gas to partial pressure of argon gas
(Given: Molar mass of N, O and Ar are 14, 16 and $40 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively.)
Arrange the following gases in increasing order of van der Waals constant 'a'
A. Ar
B. $\mathrm{CH}_{4}$
C. $\mathrm{H}_{2} \mathrm{O}$
D. $\mathrm{C}_{6} \mathrm{H}_{6}$
Choose the correct option from the following.
For 1 mol of gas, the plot of pV vs. p is shown below. p is the pressure and V is the volume of the gas
What is the value of compressibility factor at point A ?
A certain quantity of real gas occupies a volume of $0.15~ \mathrm{dm}^{3}$ at $100 \mathrm{~atm}$ and $500 \mathrm{~K}$ when its compressibility factor is 1.07 . Its volume at 300 atm and $300 \mathrm{~K}$ (When its compressibility factor is 1.4 ) is ___________ $\times 10^{-4} ~\mathrm{dm}^{3}$ (Nearest integer)
Explanation:
At $600 \mathrm{~K}$, the root mean square (rms) speed of gas $\mathrm{X}$ (molar mass $=40$ ) is equal to the most probable speed of gas $\mathrm{Y}$ at $90 \mathrm{~K}$. The molar mass of the gas $\mathrm{Y}$ is ___________ $\mathrm{g} ~\mathrm{mol}^{-1}$. (Nearest integer)
Explanation:
The root mean square speed ($v_{rms}$) of a gas can be calculated using the following formula:
$v_{rms} = \sqrt{\frac{3kT}{M}}$
where:
- $k$ is Boltzmann's constant ($1.38 \times 10^{-23}$ J/K),
- $T$ is the temperature in Kelvin,
- $M$ is the molar mass of the gas in kilograms per mole.
The most probable speed ($v_{mp}$) of a gas can be calculated using the following formula:
$v_{mp} = \sqrt{\frac{2kT}{M}}$
Given that the root mean square speed of gas X at 600 K is equal to the most probable speed of gas Y at 90 K, we can set the two equations equal to each other and solve for the molar mass of gas Y:
$\sqrt{\frac{3kT_X}{M_X}} = \sqrt{\frac{2kT_Y}{M_Y}}$
Square both sides to eliminate the square root:
$\frac{3kT_X}{M_X} = \frac{2kT_Y}{M_Y}$
We are asked to find $M_Y$, so let's rearrange the equation to solve for $M_Y$:
$M_Y = \frac{2kT_YM_X}{3kT_X}$
We know that $T_X = 600$ K, $M_X = 40$ g/mol $= 40 \times 10^{-3}$ kg/mol (converted from grams to kilograms), $T_Y = 90$ K, and $k = 1.38 \times 10^{-23}$ J/K. Substituting these values into the equation gives:
$M_Y = \frac{2 \times 1.38 \times 10^{-23} \text{J/K} \times 90 \text{K} \times 40 \times 10^{-3} \text{kg/mol}}{3 \times 1.38 \times 10^{-23} \text{J/K} \times 600 \text{K}}$
Solving this equation gives:
$M_Y = 0.004 \text{kg/mol}$
To convert this value back into grams per mole, multiply by 1000:
$M_Y = 4 \text{g/mol}$
So, the molar mass of gas Y is approximately 4 g/mol.
At constant temperature, a gas is at a pressure of 940.3 mm Hg. The pressure at which its volume decreases by 40% is __________ mm Hg. (Nearest integer)
Explanation:
To find the pressure at which the volume decreases by 40%, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
Let's assume the initial volume of the gas is V. According to Boyle's Law:
P₁V₁ = P₂V₂
Where P₁ is the initial pressure (940.3 mm Hg), V₁ is the initial volume, P₂ is the final pressure (which we need to find), and V₂ is the final volume (which is 40% less than the initial volume, or 0.6V).
Plugging in the values, we have:
940.3 mm Hg × V₁ = P₂ × (0.6V)
Now we can solve for P₂:
P₂ = (940.3 mm Hg × V₁) / (0.6V)
Since the question does not provide the initial volume (V₁), we cannot calculate the exact pressure. However, we can determine the pressure ratio when the volume decreases by 40%.
P₂ = (940.3 mm Hg × V₁) / (0.6V)
We can simplify this by canceling out V:
P₂ = 940.3 mm Hg / 0.6
P₂ ≈ 1567 mm Hg (nearest integer)
Therefore, the pressure at which the volume decreases by 40% is approximately 1567 mm Hg.
Three bulbs are filled with $\mathrm{CH}_{4}, \mathrm{CO}_{2}$ and $\mathrm{Ne}$ as shown in the picture. The bulbs are connected through pipes of zero volume. When the stopcocks are opened and the temperature is kept constant throughout, the pressure of the system is found to be ___________ atm. (Nearest integer)
Explanation:
From Dalton's partial pressure law,
$ \begin{aligned} & P_f V_f=P_1 V_1+P_2 V_2+P_3 V_3 \\\\ & P_f \times 9=2 \times 2+4 \times 3+3 \times 4 \\\\ & P_f=\frac{28}{9}=3.11=3 \end{aligned} $
The total pressure of a mixture of non-reacting gases $\mathrm{X}(0.6 \mathrm{~g})$ and $\mathrm{Y}(0.45 \mathrm{~g})$ in a vessel is $740 \mathrm{~mm}$ of $\mathrm{Hg}$.
The partial pressure of the gas $\mathrm{X}$ is _______ $\mathrm{mm}$ of $\mathrm{Hg}$. (Nearest Integer)
(Given : molar mass $\mathrm{X}=20$ and $\mathrm{Y}=45 \mathrm{~g} \mathrm{~mol}^{-1}$)
Explanation:
A $300 \mathrm{~mL}$ bottle of soft drink has $0.2 ~\mathrm{M}~ \mathrm{CO}_{2}$ dissolved in it. Assuming $\mathrm{CO}_{2}$ behaves as an ideal gas, the volume of the dissolved $\mathrm{CO}_{2}$ at $\mathrm{STP}$ is ____________ $\mathrm{mL}$. (Nearest integer)
Given : At STP, molar volume of an ideal gas is $22.7 \mathrm{~L} \mathrm{~mol}^{-1}$
Explanation:
Moles = 0.3 $\times$ 0.2
Volume at STP = $0.3\times0.2\times22.7$
= 1.362 litre
= 1362 mL
The number of statement/s, which are correct with respect to the compression of carbon dioxide from point (a) in the Andrews isotherm from the following is ___________
A. Carbon dioxide remains as a gas upto point (b)
B. Liquid carbon dioxide appears at point (c)
C. Liquid and gaseous carbon dioxide coexist between points (b) and (c)
D. As the volume decreases from (b) and (c), the amount of liquid decreases
Explanation:
(a) $\rightarrow \mathrm{CO}_2$ exist as gas
(b) $\rightarrow$ liquefaction of $\mathrm{CO}_2$ starts
(c) $\rightarrow$ liquefaction ends
(d) $\rightarrow \mathrm{CO}_2$ exist as liquid.
Between (b) & (c) $\rightarrow$ liquid and gaseous $\mathrm{CO}_2$ co-exist. As volume changes from (b) to (c) gas decreases and liquid increases.
(A), (C) $\rightarrow$ correct
Which amongst the given plots is the correct plot for pressure (p) vs density (d) for an ideal gas?
An evacuated glass vessel weighs 40.0 g when empty, 135.0 g when filled with a liquid of density 0.95 g mL$-$1 and 40.5 g when filled with an ideal gas at 0.82 atm at 250 K. The molar mass of the gas in g mol$-$1 is :
(Given : R = 0.082 L atm K$-$1 mol$-$1)
'x' g of molecular oxygen $\left(\mathrm{O}_{2}\right)$ is mixed with $200 \mathrm{~g}$ of neon (Ne). The total pressure of the non-reactive mixture of $\mathrm{O}_{2}$ and Ne in the cylinder is 25 bar. The partial pressure of Ne is 20 bar at the same temperature and volume. The value of 'x' is _________.
[Given: Molar mass of $\mathrm{O}_{2}=32 \mathrm{~g} \mathrm{~mol}^{-1}$.
Molar mass of $\mathrm{Ne}=20 \mathrm{~g} \mathrm{~mol}^{-1}$]
Explanation:
$20=25\left[X_{N e}\right]$
${\left[X_{N e}\right]=\frac{4}{5} }$
$\Rightarrow\left[\frac{\frac{200}{20}}{\frac{200}{20}+\frac{x}{32}}\right]=\frac{4}{5}$
$ \Rightarrow $$\frac{10}{10+\left(\frac{x}{32}\right)}=\frac{4}{5}$
$ \Rightarrow $$50=40+\frac{x}{8}$
$ \Rightarrow $$400=320+x$
$ \Rightarrow $$x=80$ gram
For a real gas at $25^{\circ} \mathrm{C}$ temperature and high pressure (99 bar) the value of compressibility factor is 2, so the value of Vander Waal's constant 'b' should be __________ $\times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1}$ (Nearest integer)
(Given $\mathrm{R}=0.083 \mathrm{~L}$ bar $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$)
Explanation:
$P(V-b)=R T$
$\mathrm{PV}-\mathrm{Pb}=\mathrm{RT}$
$\frac{\mathrm{PV}}{\mathrm{RT}}=1+\frac{\mathrm{Pb}}{\mathrm{RT}}$
$Z=1+\frac{\mathrm{Pb}}{\mathrm{RT}}$
$1=\frac{99(\mathrm{~b})}{0.083 \times 298}$
$b=\frac{0.083 \times 298}{99} \simeq 0.249 \simeq 25 \times 10^{-2}$
A $10 \mathrm{~g}$ mixture of hydrogen and helium is contained in a vessel of capacity $0.0125 \mathrm{~m}^{3}$ at 6 bar and $27^{\circ} \mathrm{C}$. The mass of helium in the mixture is ____________ g. (nearest integer)
Given: $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
(Atomic masses of $\mathrm{H}$ and $\mathrm{He}$ are $1 \mathrm{u}$ and $4 \mathrm{u}$, respectively)
Explanation:
$ \begin{aligned} &=\frac{P V}{R T} \\ &=\frac{6 \times 10^{5} \times 0.0125}{8.3 \times 300}=3 \end{aligned} $
Let the mass of He in $10 \mathrm{~g}$ mixture be $\mathrm{xg}$
$\therefore \frac{x}{4}+\frac{10-x}{2}=3$
On solving $x=8 \mathrm{~g}$
$\therefore$ Mass of $\mathrm{He}$ in the mixture $=8 \mathrm{~g}$
A mixture of hydrogen and oxygen contains $40 \%$ hydrogen by mass when the pressure is $2.2$ bar. The partial pressure of hydrogen is bar. (Nearest Integer)
Explanation:
Wt. of $\mathrm{H}_{2}=40 \mathrm{~g}$
Wt. of $\mathrm{O}_{2}=60 \mathrm{~g}$
$\chi_{\mathrm{H}_{2}}=\frac{\mathrm{n}_{\mathrm{H}_{2}}}{\mathrm{n}_{\mathrm{H}_{2}}+\mathrm{n}_{\mathrm{O}_{2}}}$
$=\frac{\frac{40}{2}}{\frac{40}{2}+\frac{60}{32}}$
$=\frac{20}{20+1.875}$
$=\frac{20}{21.875}=0.914$
$\mathrm{P}_{\mathrm{H}_{2}}=\chi_{\mathrm{H}_{2}} \times \mathrm{P}_{\mathrm{T}}$
$=0.914 \times 2.2$
$=2.01 \simeq 2 \mathrm{bar}$
A sealed flask with a capacity of 2 dm3 contains 11 g of propane gas. The flask is so weak that it will burst if the pressure becomes 2 MPa. The minimum temperature at which the flask will burst is ___________ $^\circ$C. [Nearest integer]
(Given : R = 8.3 J K$-$1 mol$-$1 , Atomic masses of C and H are 12u and 1u, respectively.) (Assume that propane behaves as an ideal gas.)
Explanation:
$\mathrm{PV}=\mathrm{nRT}$
$P=2 \times 10^{6} \mathrm{~Pa}$
$V=2 \mathrm{dm}^{3}=2 \times 10^{-3} \mathrm{~m}^{3}$
$\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{n}=\frac{11}{44} \mathrm{~mol}$
$2 \times 10^{6} \times 2 \times 10^{-3}=\frac{11}{44} \times 8.3 \times \mathrm{T}$
$\mathrm{T}=1927.7 \mathrm{~K}$
$\mathrm{T}\left(\right.$ in $\left.^{\circ} \mathrm{C}\right)=1927.7-273 \simeq 1655^{\circ} \mathrm{C}$
The pressure of a moist gas at $27^{\circ} \mathrm{C}$ is $4 \mathrm{~atm}$. The volume of the container is doubled at the same temperature. The new pressure of the moist gas is ________________ $\times 10^{-1} \mathrm{~atm}$. (Nearest integer)
(Given : The vapour pressure of water at $27^{\circ} \mathrm{C}$ is $0.4 \mathrm{~atm} .$ )
Explanation:
To solve this problem, we need to determine the new pressure of the moist gas when the volume of the container is doubled while maintaining the same temperature. First, we need to separate the contributions of the gas and the water vapor to the initial pressure and then apply Boyle's Law for the gas component only. The steps are as follows:
1. Initial pressure of the moist gas (gas + water vapor) is given as $4 \, \text{atm}$.
The vapour pressure of water at $27^{\circ} \mathrm{C}$ is $0.4 \, \text{atm}$.
2. The partial pressure of the dry gas component can be calculated by subtracting the vapor pressure of water from the initial pressure:
$ P_{\text{dry gas (initial)}} = 4 \, \text{atm} - 0.4 \, \text{atm} = 3.6 \, \text{atm} $
3. When the volume of the container is doubled, the pressure of the dry gas component will change according to Boyle's Law, which states:
$ P_1 V_1 = P_2 V_2 $
Since the temperature remains constant and the volume is doubled (i.e., $V_2 = 2V_1$), we can find the new pressure of the dry gas ($P_{\text{dry gas (new)}}$) with the equation:
$ 3.6 \, \text{atm} \times V_1 = P_{\text{dry gas (new)}} \times 2V_1 $
Solving for $P_{\text{dry gas (new)}}$:
$ P_{\text{dry gas (new)}} = \frac{3.6 \, \text{atm}}{2} = 1.8 \, \text{atm} $
4. The vapor pressure of water remains unchanged at $0.4 \, \text{atm}$ because it depends only on the temperature, not on the volume.
5. The new total pressure of the moist gas is the sum of the new pressure of the dry gas and the unchanged vapor pressure of water:
$ P_{\text{moist gas (new)}} = P_{\text{dry gas (new)}} + P_{\text{water vapor}} = 1.8 \, \text{atm} + 0.4 \, \text{atm} = 2.2 \, \text{atm} $
6. Finally, as requested, we express the new pressure in the form of $\times 10^{-1} \, \text{atm}$ and find the nearest integer:
$ 2.2 \, \text{atm} = 22 \times 10^{-1} \, \text{atm} $
Therefore, the nearest integer is 22.
Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 gL$-$1 at 257$^\circ$C and 100 mm Hg. The molar mass of geraniol is ____________ g mol$-$1. (Nearest Integer)
[Given : R = 0.082 L atm K$-$1 mol$-$1]
Explanation:
From ideal gas equation we know
PV = nRT
$ \Rightarrow PV = {W \over M}RT$
$ \Rightarrow P = {W \over V}\,.\,{{RT} \over M}$
$ \Rightarrow P = d\,.\,{{RT} \over M}$ [$\because$ $d = {W \over V}$]
We know, 760 mm of Hg = 1 atm
$\therefore$ 100 mm of Hg = ${{100} \over {760}}$ atm
$\therefore$ Pressure (P) = ${{100} \over {760}}$ atm
Density (d) = 0.46
R = 0.082 L atm K$-$1 mol$-$1
T = (257 + 273) K = 530 K
Putting the values in above equation, we get
${{100} \over {760}} = {{0.46 \times 0.082 \times 530} \over M}$
$\Rightarrow$ M = 152
100 g of an ideal gas is kept in a cylinder of 416 L volume at 27$^\circ$C under 1.5 bar pressure. The molar mass of the gas is __________ g mol$-$1. (Nearest integer)
(Given : R = 0.083 L bar K$-$1 mol$-$1)
Explanation:
Given, Mass of ideal gas = 100 gm
Let the molar mass of ideal gas = M
$\therefore$ Number of moles of gas (n) = ${{100} \over M}$
Volume of cylinder (V) = 416 L
Temperature (T) = (27 + 273)K = 300 K
Pressure (P) = 1.5 bar
R = 0.083 L bar K$-$1 mol$-$1
Using ideal gas equation,
PV = nRT
$ \Rightarrow 1.5 \times 416 = {{100} \over M} \times 0.083 \times 300$
$\Rightarrow$ M = 4
2.0 g of H2 gas is adsorbed on 2.5 g of platinum powder at 300 K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is __________ mL.
(Given : R = 0.083 L bar K$-$1 mol$-$1)
Explanation:
$ \mathrm{V}=\frac{2 \times 0.083 \times 300}{2 \times 1}=24.9 \text { litre } $
$\therefore$ Volume of the gas adsorbed per gram of the adsorbent
$ \begin{aligned} &=\frac{24.9}{2.5}=9.96 \mathrm{~L} \\\\ &=9960 ~\mathrm{ml} \end{aligned} $
A rigid nitrogen tank stored inside a laboratory has a pressure of 30 atm at 06:00 am when the temperature is 27$^\circ$C. At 03:00 pm, when the temperature is 45$^\circ$, the pressure in the tank will be _________ atm. [nearest integer]
Explanation:
A nitrogen tank of fixed volume used where number of moles of nitrogen is fixed.
$\therefore$ V = constant
n = constant
R = constant
From ideal gas equation,
PV = nRT
$\Rightarrow$ P $\propto$ T [As V, n, R = constant]
Here, initially P1 = 30 atm, T1 = 300 K
Finally, P2 = ?, T2 = 318 K
$\therefore$ ${{{P_1}} \over {{P_2}}} = {{{T_1}} \over {{T_2}}}$
$ \Rightarrow {{30} \over {{P_2}}} = {{300} \over {318}}$
$ \Rightarrow {P_2} = 31.8 \simeq 32$
At 300 K, a sample of 3.0 g of gas A occupies the same volume as 0.2 g of hydrogen at 200 K at the same pressure. The molar mass of gas A is ____________ g mol$-$1. (nearest integer) Assume that the behaviour of gases as ideal.
(Given : The molar mass of hydrogen (H2) gas is 2.0 g mol$-$1.)
Explanation:
Both gas A and Hydrogen (H2) gas have same volume at same pressure. Let both 's volume is V and pressure P.
For gas A :
Pressure = P
Temperature (T) = 300 K
Volume = V
Mass = 3 g
Molar mass = M gm/mol
using ideal gas equation,
PV = nRT
$ \Rightarrow PV = {3 \over M} \times R \times 300$ ..... (1)
For Hydrogen,
Pressure = P
Temperature (T) = 200 K
Volume = V
Mass = 0.2 g
Molar mass = 2 gm/mol
Using ideal gas equation,
$PV = {{0.2} \over 2} \times R \times 200$ ...... (2)
From (1) and (2), we get
${3 \over M} \times R \times 300 = {{0.2} \over 2} \times R \times 200$
$\Rightarrow$ M = 45
(Assume LPG of be an ideal gas)
Explanation:
Weight of full LPG cylinder = 29 kg
$\therefore$ Weight of gas = 29 $-$ 14.8 = 14.2 kg
If weight of full LPG cylinder = 23 kg
then weight of gas used = 29 $-$ 23 = 6 kg at ambient temperature.
From ideal gas equation, pV = nRT
or $pV = {{Weight\,of\,solute} \over {Molecular\,mass\,of\,solute}} \times RT$
or $pV = {W \over M} \times RT$
Applying ideal gas to LPG cylinder when gas is full,
$pV = nRT$
$3.47\,atm \times V = {{14.2kg} \over M} \times RT$ .... (i)
Applying ideal gas to LPG cylinder when gas is reduced to 23 kg at ambient temperture,
$pV = nRT$
$p \times V = {{8.2kg} \over M} \times RT$ .... (ii)
Divide Eq. (i) by (ii)
${{3.47 \times V} \over {p \times V}} = {{{{14.2kg\,RT} \over M}} \over {{{8.2kg\, \times RT} \over M}}}$
${{3.47} \over p} = {{14.2} \over {8.2}}$
$ \Rightarrow p = {{3.47 \times 41} \over {71}} = 2.003$ atm
Hence, answer is 2.
Explanation:
$ \Rightarrow {T_2} = 1200$ K
$ \Rightarrow $ ${T_2} = 927^\circ $ C
(Given R = 0.083 L atm K$-$1 mol$-$1)
Explanation:
$ = {{1 \times 4 \times {{10}^3} \times 1000} \over {0.083 \times 300}}$
Weight of CH4
$ = {{40 \times 16 \times {{10}^5}} \over {0.083 \times 300}}$ gm
$ = 25.7 \times {10^5}$ gm
[Assume chlorine is an ideal gas at STP
R = 0.083 L bar mol$-$1 K$-$1, NA = 6.023 $\times$ 1023]
Explanation:
$ = {{1 \times 20 \times {{10}^{ - 3}}} \over {0.083 \times 273}}$
No. of atoms = $ = {{1 \times 20 \times {{10}^{ - 3}}} \over {0.083 \times 273}} \times 2 \times 6.023 \times {10^{23}}$
$ = 1.06 \times {10^{21}}$
[Assume gases are ideal, R = 8.314 J mol$-$1 K$-$1
Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u]
Explanation:
(m)methane = 6.4 g, (m)CO2 = 8.8 g
PV = ntotalRT
P $\times$ 10 $\times$ 10$-$3 = $\left( {{{6.4} \over {16}} + {{8.8} \over {44}}} \right)$ $\times$ 8.314 $\times$ 300
P $\times$ 10$-$2 = (0.4 + 0.2) $\times$ 8.314 $\times$ 300
P = 149652 Pa
P = 149.652 KPa $ \approx $ 150 kPa
[R = 0.0821 L atm K$-$1mol$-$1]
Explanation:
Using ideal gas equation : PV = nRT
V = ${{0.0975 \times 0.082 \times 300} \over 1}$ = 2.4 L
$ \therefore $ Volume of O2(g) adsorbed per gram of the
adsorbent = ${{2.4} \over {1.2}}$ = 2
Explanation:
For 1 mole of a real gas, the van der Waals' equation is,
$\left( {p + {a \over {V_m^2}}} \right)({V_m} - b) = RT$
At very high pressure, the equation becomes,
$p({V_m} - b) = RT$
$ \Rightarrow p{V_m} = RT + pb \Rightarrow {{p{V_m}} \over {RT}} = 1 + {{pb} \over {RT}}$
$ \Rightarrow Z = 1 + {{pb} \over {RT}}$ [$\because$ $Z = {{p{V_m}} \over {RT}}$ = compressibility]
$\therefore$ ${\left( {{{\delta Z} \over {\delta p}}} \right)_T} = 0 + {b \over {RT}} + {b \over {RT}} = {{xb} \over {RT}}$ (Given)
$ \Rightarrow x = 1$
Explanation:
${{35} \over {300}} = {{40} \over {{T_2}}}$
${T_2} = {{40 \times 300} \over {35}}$
$ = 342.86$ K
$ = 69.85^\circ $ C $ \simeq 70^\circ $ C
[Given R = 0.0826 L atm K$-$1 mol$-$1]
Explanation:
Molecular weight = 26 g/mol
Temperature = 50 + 273 = 323 K
Pressure = 740 torr/mm of Hg
Pressure = ${{740} \over {760}}$ atm
R = 0.0821 L atm mol$-$1 K$-$1
Hence, no. of mole n = ${{4.75} \over {26}}$ mol
Formula used, pV = nRT (ideal gas)
$ \Rightarrow V = {{nRT} \over p} = {{4.75} \over {26}} \times {{0.0821 \times 323} \over {(740/760)}}$
$ = {{96314.078} \over {19240}} = 5.0059$ L = 5 L
d = Density, P = Pressure, T = Temperature
Root mean square speed (Vrms); most probable speed (Vmp); Average speed (Vav)
a pressure of 48 $ \times $ 10–3 bar. At the same temperature, the pressure, of a spherical balloon of radius 12 cm containing the
same amount of gas will be ________ $ \times $ 10–6 bar.
Explanation:
According to Boyle's law,
P1V1 = P2V2
Given, P1 = 48 $ \times $ 10–3 bar
V1 = ${{4 \over 3}\pi {{\left( 3 \right)}^3}}$
V2 = ${{4 \over 3}\pi {{\left( 12 \right)}^3}}$
P2 = ?
$ \therefore $ P2 = ${{{P_1}{V_1}} \over {{V_2}}}$
= ${{48 \times {{10}^{ - 3}} \times {{\left( 3 \right)}^3}} \over {{{\left( {12} \right)}^3}}}$
= 7.5 $ \times $ 10-4 = 750 $ \times $ 10-6 bar
NaClO3(s) + Fe(s) $ \to $ O2(g) + NaCl(s) + FeO(s)
R = 0.082 L atm mol–1 K–1
Explanation:
moles of NaClO3 = moles of O2
moles of O2 = ${{PV} \over {RT}}$ = ${{1 \times 492} \over {0.082 \times 300}}$ = 20 mol
mass of NaClO3 = 20 $ \times $ 106.5 = 2130 g
(Vmp : most probable velocity)
| Gas | a/(k Pa dm6 mol-1) | b/(dm3 mol-1) |
|---|---|---|
| A | 642.32 | 0.05196 |
| B | 155.21 | 0.04136 |
| C | 431.91 | 0.05196 |
| D | 155.21 | 0.4382 |
a and b are vander Waals constants. The correct statement about the gases is :
$p = {{RT} \over {V - b}}$ at T.
Here, b is the van der Waals constant. Which gas will exhibit steepest increase in the plot of Z (compression factor) vs p?
| Gas | Ar | Ne | Kr | Xe |
|---|---|---|---|---|
| a/ (atm dm6 mol–2) | 1.3 | 0.2 | 5.1 | 4.1 |
| b/ (10–2 dm3 mol–1) | 3.2 | 1.7 | 1.0 | 5.0 |
Which gas is expected to have the highest critical temperature?