Ideal Gas Equation from KTG
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The root mean square (R.M.S.) Speed V of the molecules of an ideal gas is given by the expressions,
v = √ (3RT/M) and
v = √ (3KT/m) where R is universal gas constant, T is the absolute (Kelvin) temperature M is the molar mass, K is Boltzman's constant and M is the molecular mass. The R.M.S. speed of oxygen molecules (O2) at temperature T1 is V1. When the temperature is doubled, if the oxygen molecules are dissociated into atomic oxygen, what will be R.M.S. speed of oxygen atoms? (Treat the gas as ideal).
v1/2
v1
2v1
One mole of ideal gas follows a law according to that it's pressure varies with volume as p= a1+vbwhere a and b are positive constant the temperature of the gas at volume v= a is
A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at a temperature 2T has a pressure of
P8
P
2P
8P
4 moles of an ideal gas is at 0∘C. At constant pressure it is heated to double its volume, then its final temperature will be
0°C
273°C
546°C
136.5°C
A long cylindrical tank of cross-section area 0.5m2 is filled with water. It has an opening at a height 50 cm from the bottom, having area of cross-section 1×10−4m2 . A movable piston of cross-section area almost equal to 0.5 m2 is fitted on the top of the tank such that it can slide in the tank freely. A load of 20 kg is applied on the top of the water by piston, as shown in the figure. Find the speed of the water jet with which it hits the surface when piston is 1m above the bottom (Ignore the mass of the piston).
3.56 m/s
3.3 m/s
5.56 m/s
4.56 m/s
- Celsius constant
- Characteristic gas constant
- Universal gas constant
- Boltzmann’s constant
A quantity of air is taken from state a to state b along a path that is a straight line in the P-V diagram.In this process, what happens to the temperature of the gas?
Decreases
Remains constant
Not enough information
Increases
- 67.2 g
- 40 g
- 20.4 g
- 10 g
Two thermally insulated vessels 1 and 2 are filled with air at temperature (T1, T2), volume (V1, V2) and pressure (P1, P2) respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be
T1 +T2
(T1 + T2)/2
T1T2(P1V1+P2V2)/P1V1T2+P2V2T1)
T1T2(P1V1+P2V2)/P1VT1+P2V2T2)
- 1T
- 2T
- 3T
- 4T
- 2.5×105 Pa
- 2.0×105 Pa
- 3.0×105 Pa
- 1.5×105 Pa
Two thermally insulated vessels 1 and 2 are filled with air at temperature (T1, T2), volume (V1, V2) and pressure (P1, P2) respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be
T1 + T2
T1 + T22
T1T2(P1V1+P2V2)P1V1T2+P2V2T1)
T1T2(P1V1+P2V2)P1V1T1+P2V2T2)
An enclosure of volume V contains a mixture of 8 g of oxygen, 14 g of nitrogen and 22 g of carbon- dioxide at absolute temperature T. The pressure of the mixture of gases is (R is universal gas constant)