Ideal Gas Equation from KTG
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What is the unit of Van der Waals constant?
The Ideal Gas Law is applicable at
Low , Low
High , Low
Low , High
High , High
A vessel of volume V contains an ideal gas at absolute temperature T and pressure P. The gas is allowed to leak till its pressure falls to P. Assuming that the temperature remains constant during leakage, the number of moles of the gas that have leaked is
VRT(P + P′)
V2RT(P + P′)
VRT(P − P′)
V2RT(P − P′)
- 1T
- 2T
- 3T
- 4T
- True
- False
- mass of the gas
- kinetic energy of the gas
- number of moles of the gas
- number of molecules in the gas
- 1T
- 2T
- 3T
- 4T
- R
- 10R
- 20R
- 30R
- 2.5×105 Pa
- 2.0×105 Pa
- 3.0×105 Pa
- 1.5×105 Pa
What is ideal gas law used for?
- p=p0e−(MgRT)h
- p=p0e(MgRT)h2
- p=p0e−(MgRT)h2
- p=p0e−(MgRT)h
- 1
- 1.33
- 1.67
- 0.67
- P=43P0
- P=23P0
- n1=23P0V0RT0
- n2=43P0V0RT0
- P0A2μr
- P0Aμπr
- P0A4μπr
- P0A2μπr
- T
- 9T
- 27T
- T9
- True
- False
- 2.5×105 Pa
- 2.0×105 Pa
- 3.0×105 Pa
- 1.5×105 Pa
Air is pumped into an automobile tyre's tube up to a pressure of 200 kPa in the morning when the air temperature is 200∘C. During the day the temperature rises to 400∘C and the tube expands by 2%. Calculate the pressure of the air in the tube at this temperature
209 kPa
285 kPa
355 kPa
392 kPa
- P/(kTV)
- mkT
- P/(kT)
- Pm/(kT)
A vessel of volume V contains an ideal gas at absolute temperature T and pressure P. The gas is allowed to leak till its pressure falls to P. Assuming that the temperature remains constant during leakage, the number of moles of the gas that have leaked is
VRT(P + P′)
V2RT(P + P′)
VRT(P − P′)
V2RT(P − P′)
- T
- 9T
- 27T
- T9
- 1T
- 2T
- 3T
- 4T
[Given: Molecular weight of air = 29, 1 atm = 1.013 ×105 Pa]
- 105.85 kg
- 29 kg
- 74.37 kg
- 135.35 kg
560 cm3 of air is at a pressure of 76 cm of mercury. Find the pressure of air (assuming temperature remains constant), when its volume is:
840 cm3
Air is pumped into an automobile tyre's tube up to a pressure of 200 kPa in the morning when the air temperature is 200∘C. During the day the temperature rises to 400∘C and the tube expands by 2%. Calculate the pressure of the air in the tube at this temperature
209 kPa
285 kPa
355 kPa
392 kPa
- 0.25 atm
- 0.5 atm
- 4 atm
- 1T
- 2T
- 3T
- 4T
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)
- 16:1
- 1:16
- 1:1
- 1:8
- 1.25×105 Pa
- 2.25×105 Pa
- 3.25×105 Pa
- 4.25×105 Pa