Advertisements
Advertisements
Question
Calculate the volume of 1 mole of an ideal gas at STP.
Solution
Here,
STP means a system having a temperature of 273 K and 1 atm pressure.
Pressure, P = 1.01325\[\times\]05 Pa
No of moles, n = 1 mol
Temperature, T = 273 K
Applying the equation of an ideal gas, we get
PV = nRT
⇒ V =\[\frac{RT}{P}\]
⇒ V=\[\frac{8 . 314 \times 273}{1 . 01325 \times {10}^5} = 0 . 0224 \text{ m}^3\]
APPEARS IN
RELATED QUESTIONS
The average momentum of a molecule in a sample of an ideal gas depends on
Consider the quantity \[\frac{MkT}{pV}\] of an ideal gas where M is the mass of the gas. It depends on the
A sample of 0.177 g of an ideal gas occupies 1000 cm3 at STP. Calculate the rms speed of the gas molecules.
Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an adiabatic process.
(a) Q = 0
(b) W = 0
(c) Q = W
(d) Q ≠ W
A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperature of the gas increases by 1° C if 3.0 cal of heat is added to it. The gas may be
(a) helium
(b) argon
(c) oxygen
(d) carbon dioxide
An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation p = kV. Show that the molar heat capacity of the gas for the process is given by `"C" ="C"_"v" +"R"/2.`
An ideal gas (Cp / Cv = γ) is taken through a process in which the pressure and the volume vary as p = aVb. Find the value of b for which the specific heat capacity in the process is zero.
Half mole of an ideal gas (γ = 5/3) is taken through the cycle abcda, as shown in the figure. Take `"R" = 25/3"J""K"^-1 "mol"^-1 `. (a) Find the temperature of the gas in the states a, b, c and d. (b) Find the amount of heat supplied in the processes ab and bc. (c) Find the amount of heat liberated in the processes cd and da.
An ideal gas (γ = 1.67) is taken through the process abc shown in the figure. The temperature at point a is 300 K. Calculate (a) the temperatures at b and c (b) the work done in the process (c) the amount of heat supplied in the path ab and in the path bcand (d) the change in the internal energy of the gas in the process.
Consider a given sample of an ideal gas (Cp/Cv = γ) having initial pressure p0 and volume V0. (a) The gas is isothermally taken to a pressure p0/2 and from there, adiabatically to a pressure p0/4. Find the final volume. (b) The gas is brought back to its initial state. It is adiabatically taken to a pressure p0/2 and from there, isothermally to a pressure p0/4. Find the final volume.
Two samples A and B, of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 − 21−γ = (γ − 1) ln2.
1 litre of an ideal gas (γ = 1.5) at 300 K is suddenly compressed to half its original volume. (a) Find the ratio of the final pressure to the initial pressure. (b) If the original pressure is 100 kPa, find the work done by the gas in the process. (c) What is the change in internal energy? (d) What is the final temperature? (e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. (f) The gas is now expanded isothermally to achieve its original volume of 1 litre. Calculate the work done by the gas. (g) Calculate the total work done in the cycle.
Figure shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas (γ = 1.5) is injected in the two sides at equal pressures and temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio 1 : 3. Find the ratio of the temperatures in the two parts of the vessel.
Two vessels A and B of equal volume V0 are connected by a narrow tube that can be closed by a valve. The vessels are fitted with pistons that can be moved to change the volumes. Initially, the valve is open and the vessels contain an ideal gas (Cp/Cv = γ) at atmospheric pressure p0 and atmospheric temperature T0. The walls of vessel A are diathermic and those of B are adiabatic. The valve is now closed and the pistons are slowly pulled out to increase the volumes of the vessels to double the original value. (a) Find the temperatures and pressures in the two vessels. (b) The valve is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of the temperature and pressure.
The figure shows an adiabatic cylindrical tube of volume V0 divided in two parts by a frictionless adiabatic separator. Initially, the separator is kept in the middle, an ideal gas at pressure p1 and temperature T1 is injected into the left part and another ideal gas at pressure p2 and temperature T2 is injected into the right part. Cp/Cv = γ is the same for both the gases. The separator is slid slowly and is released at a position where it can stay in equilibrium. Find (a) the volumes of the two parts (b) the heat given to the gas in the left part and (c) the final common pressure of the gases.
An ideal gas of density 1.7 × 10−3 g cm−3 at a pressure of 1.5 × 105 Pa is filled in a Kundt's tube. When the gas is resonated at a frequency of 3.0 kHz, nodes are formed at a separation of 6.0 cm. Calculate the molar heat capacities Cp and Cv of the gas.
1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K (Figure). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time ______.
ABCDEFGH is a hollow cube made of an insulator (Figure). Face ABCD has positive charge on it. Inside the cube, we have ionized hydrogen. The usual kinetic theory expression for pressure ______.
- will be valid.
- will not be valid since the ions would experience forces other than due to collisions with the walls.
- will not be valid since collisions with walls would not be elastic.
- will not be valid because isotropy is lost.
