1 Crore+ students have signed up on EduRev. Have you? 
Q. 1. A vessel of volume V = 30 l contains ideal gas at the temperature 0°C. After a portion of the gas has been let out, the pressure in the vessel decreased by Δp = 0.78 atm (the temperature remaining constant). Find the mass of the released gas. The gas density under the normal conditions p = 1.3 g/I.
Solution. 1. Let m_{1 }and m_{2} be the masses of the gas in the vessel before and after the gas is released. Hence mass of the gas released,
Now from ideal gas equation
as V and T are same before and after the release of the gas.
so,
or, (1)
We also know (2)
(where p_{0} = standard atmospheric pressure and
From Eqs. (1) and (2) we get
Q. 2. Two identical vessels are connected by a tube with a valve letting the gas pass from one vessel into the other if the pressure difference atm. Initially there was a vacuum in one vessel while the other contained ideal gas at a temperature t_{1} = 27°C and pressure p_{1} = 1.00 atm. Then both vessels were heated to a temperature t_{2} =107°C. Up to what value will the pressure in the first vessel (which had vacuum initially) increase?
Solution. 2. Let m_{1} be the mass of the gas enclosed.
Then,
When heated, some gas, passes into the evacuated vessel till pressure difference becomes be the pressure on the two sides of the valve. Then and
But,
So,
or,
Q. 3. A vessel of volume V = 20 l contains a mixture of hydrogen and helium at a temperature t = 20°C and pressure p = 2.0 atm. The mass of the mixture is equal to m = 5.0 g. Find the ratio of the mass of hydrogen to that of helium in the given mixture.
Solution. 3. Let the mixture contain v_{1} and v_{2} moles of H_{2} and H_{e} respectively. If molecular weights of H_{2} and H_{e} are M_{1} and M_{2}, then respective masses in the mixture are equal to
m_{1} = v_{1}M_{1} and m_{2} = v_{2}M_{2}
Therefore, for the total mass of the mixture we get,
m = m_{1} + m_{2} or m = v_{1} M_{1} + v_{2} M_{2} (1)
Also, if v is the total number of moles of the mixture in the vessels, then we know,
V = V_{1} + V_{2} (2)
Solving (1) and (2) for v_{1} and v_{2}, we get,
Therefore, we get
or,
One can also express the above result in terms of the effective molecular weight M of the mixture, defined as,
Thus,
Using the data and table, we get :
Q. 4. A vessel contains a mixture of nitrogen (m_{1} = 7.0 g) and carbon dioxide (m_{2} = 11 g) at a temperature T = 290 K and pressure p_{0}, = 1.0 atm. Find the density of this mixture, assuming the gases to be ideal.
Solution. 4. We know, for the mixture, N_{2} and CO_{2} (being regarded as ideal gases, their mixture too behaves like an ideal gas)
where, v is the total number of moles of the gases (mixture) present and V is the volume of the vessel. If v_{1} and v_{2} are number of moles of N_{2} and CO_{2} respectively present in the mixture, then
v = v_{1} + v_{2}
Now number of moles of N_{2} and CO_{2} is, by definition, given by
where, m_{1} is the mass of N_{2} (Moleculer weight = M_{1}) in the mixture and m_{2} is the mass of CO_{2} (Molecular weight = M_{2}) in the mixture.
Therefore density of the mixture is given by
Q. 5. A vessel of volume V = 7.5 1 contains a mixture of ideal gases at a temperature T = 300 K: v_{1} = 0.10 mole of oxygen, v_{2} = 0.20 mole of nitrogen, and v_{3} = 0.30 mole of carbon dioxide. Assuming the gases to be ideal, find:
(a) the pressure of the mixture;
(b) the mean molar mass M of the given mixture which enters its equation of state pV = (m/M) RT, where m is the mass of the mixture.
Solution. 5. (a) The mixture contains v_{1}, v_{2} and v_{3} moles of O_{2 }> N_{2} and CO_{2} respectively. Then the total number of moles of the mixture
v = v_{1} + v_{2} + v_{3}
We know, ideal gas equation for the mixture
or,
(b ) M ass o f oxygen (O_{2}) present in the mixture : m_{1} = v_{1} M_{1}
Mass of nitrogen (N_{2}) present in the mixture : m_{2} = v_{2}M_{2}
Mass of carbon dioxide (CO_{2}) present in the mixture : m_{3} = v_{3}M_{3}
So, mass of the mixture
Moleculer mass of the mixture
Q. 6. A vertical cylinder closed from both ends is equipped with an easily moving piston dividing the volume into two parts, each containing one mole of air. In equilibrium at T_{0} = 300 K the volume of the upper part is η = 4.0 times greater than that of the lower part. At what temperature will the ratio of these volumes be equal η' = 3.0?
Solution. 6. Let p_{1} and p_{2} be the pressure in the upper and lower part of the cylinder respectively at temperature T_{0}. At the equilibrium position for the piston :
(m is the mass o f the piston.)
But (where V_{0} is the initial volume of the lower part)
So, (1)
Let T be the sought temperature and at this temperature the volume of the lower part becomes V, then according to the problem the volume of the upper part becomes η'V'
Hence, (2)
From (1) and (2).
As, the total volume must be constant,
Putting the value of V in Eq. (3), we get
Q. 7. A vessel of volume V is evacuated by means of a piston air pump. One piston stroke captures the volume ΔV. How many strokes are needed to reduce the pressure in the vessel η times? The process is assumed to be isothermal, and the gas ideal.
Solution. 7. Let P_{1} be the density after the first stroke. The the mass remains constant
Similarly, if p_{2} is the density after second stroke
In this way after nth stroke.
Since pressure α density,
(because temperature is constant)
It is required by
so,
Hence
Q. 8. Find the pressure of air in a vessel being evacuated as a function of evacuation time t. The vessel volume is V, the initial pressure is P_{0}. The process is assumed to be isothermal, and the evacuation rate equal to C and independent of pressure.
Note. The evacuation rate is the gas volume being evacuated per unit time, with that volume being measured under the gas pressure attained by that moment.
Solution. 8. From the ideal gas equation
(1)
In each stroke, volume v of the gas is ejected, where v is given by
In case of continuous ejection, if (m_{N1}) corresponds to mass of gas in the vessel at time t, then m_{N} is the mass at time t + Δt, where Δt, is the time in which volume v of the ga
has come out. The rate of evacuation is therefore
In the lim it Δt  0, we get
(2)
From (1) and (2)
Integrating
Thus
Q. 9. A chamber of volume V = 87 l is evacuated by a pump whose evacuation rate (see Note to the foregoing problem) equals C = 10 1/s. How soon will the pressure in the chamber decrease by η = 1000 times?
Solution. 9. Let p be the instantaneous density, then instantaneous mass = V_{p}. In a short interval dt the volume is increased by Cdt.
So,
(because mass remains constant in a short interval dt)
so,
Since pressure α density
or
or
Q. 10. A smooth vertical tube having two different sections is open from both ends and equipped with two pistons of different areas (Fig. 2.1). Each piston slides within a respective tube section. One mole of ideal gas is enclosed between the pistons tied with a nonstretchable thread. The crosssectional area of the upper piston is ΔS = 10 cm^{2} greater than that of the lower one. The combined mass of the two pistons is equal to m = 5.0 kg. The outside air pressure is P_{0} = 1.0 atm. By how many kelvins must the gas between the pistons be heated to shift the pistons through l = 5.0 cm?
Solution. 10. The physical system consists of one mole of gas confined in the smooth vertical tube. Let m_{1} and m_{2} be the masses of upper and lower pistons and S_{1} and S_{2} are their respective areas.
For the lower piston
or, (1)
Similarly for the upper piston
or, (2)
From (1) and (2)
or,
so,
From the gas law, p V = v R T
(because p is constant)
So,
Hence,
Q. 11. Find the maximum attainable temperature of ideal gas in each of the following processes:
where p_{0},,α and β are positive constants, and V is the volume of one mole of gas.
Solution. 11.
Thus, (1)
which yields, (2)
Hence,
so (1)
For T_{max} the condition is which yields
Hence using this value of p in Eq. (1), we get
Q. 12. Find the minimum attainable pressure of ideal gas in the process T = T_{0} + α V^{2}, where T_{0} and α are positive constants, and V is the volume of one mole of gas. Draw the approximate p vs V plot of this process.
Solution. 12.
(as, V = RT/p for one mole of gas)
So, (1)
For
T = 2 T_{0} (2)
From (1) and (2), we get,
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 