Which one of the following indicates the value of the gas constant R?
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Solution
8.31 J.K-1 mol-1
1 cal= 4.2J.
∴\(\frac{8.31}{4.2}\)cal.K-1mol-1=1.987calK-1mol-1
Van der Waal’s equation \(\left [ P+\frac{a}{V^{2}} \right ](V-b)=nRT\)is applicable for :
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Solution
Vandar Waal’s equation is applicable for real (non-ideal)gases.
Volume occupied by a gas at one atmospheric pressure and 0°C is V mL. Its volume at 273 K will be
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Solution
0°C is equivalent to 273° K i.e., conditions are same so volume will be V ml.
A cylinder of 5 L capacity, filled with air at NTP is connected with another evacuated cylinder of 30 L of capacity. There sultant air pressure in both the cylinders will
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Solution
The gas inside the cylinder of 5L capacity has pressure 76 cm of Hg (NTP)The new volume for the gas = (30 + 5)L = 35 L.
According to Boyle's law:P1V1= P2V2
76 × 5 = P2 × 35\(∴P_{2}=\frac{76\times 5}{35}=10.85\cong 10.8\: cm\: of\: Hg\)
Densities of two gases are in the ratio 1:2 and their temperatures are in the ratio 2:1 then the ratio of their respective pressures is
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Solution
P∝d and T,\(\frac{P_{1}}{P_{2}}=\frac{d_{1}T_{1}}{d_{2}T_{2}}=\frac{1}{2}\times \frac{2}{1}\Rightarrow 1:1\)
Gas equation PV= nRT is obeyed by
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Solution
PV = nRT is for an ideal gas which follows both isothermal and adiabatic processes.
The ratio of the rate of diffusion of helium and methane under identical condition of pressure and temperature will be
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Solution
Use Grahams’ law of diffusion
\(\frac{r_{He}}{r_{CH_{4}}}=\sqrt{\frac{M_{CH_{4}}}{M_{He}}}=\sqrt{\frac{16}{4}}=2\)
From a heated mixture of nitrogen, oxygen and carbon,two compounds (out of the many obtained) are isolated. Therates of diffusion of the two isolated compounds are almostidentical. The two compounds are
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Solution
Rate of diffusion depend upon molecular weight
\(\frac{r_{1}}{r_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}\Rightarrow r_{1}=r_{2}\: if\: M_{1}=M_{2}\)
Hence, compounds are N2O and CO2 as both have samemolar mass.
600 c.c. of a gas at a pressure of 750 mm of Hg is compressed to 500 c.c. Taking the temperature to remain constant, the increase inpressure, is
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Solution
Given initial volume (V1) = 600 c.c.; Initial pressure(P1) = 750 mm of Hg and final volume (V2) = 500 c.c.according to Boyle’s law,
P1V1= P2V2or 750 × 600 = P2 × 500
or P2=\(\frac{750\times 600}{500}=900\: mm\: of\: hg\)
Therefore increase of pressure = (900 – 750) = 150 mm of Hg
500 ml of nitrogen at 27°C is cooled to –5°C at the same pressure. The new volume becomes
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Solution
Given initial volume (V1) = 500 ml ; Initial temperature(T1) = 27ºC = 300 K and final temperature (T2) = –5ºC= 268 K.
From Charle’s law :\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\: or\: \frac{500}{300}=\frac{V_{2}}{268}\)
Where V2= New volume of gas
\(V_{2}=\frac{500}{300}\times 268=446.66ml\)