(6) There is a cylindrical container with a piston with diameter d = 4.67cm which allows the volume to change. Sitting on top of that piston is a m = 6.62kg block. The temperature of the diatomic gas is T 382K and there are N = 1.45 × 1023 molecules. (a) What is the gauge pressure of the gas inside the container? (b) What is the volume of this container in cm³? (c) What is the mean free path of a molecule in this container?

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(6) There is a cylindrical container with a piston with diameter d:
volume to change. Sitting on top of that piston is a m =
the diatomic gas is T
4.67cm which allows the
6.62kg block. The temperature of
382K and there are N
1.45 x 1023 molecules.
||
(a) What is the gauge pressure of the gas inside the container?
(b) What is the volume of this container in cm³?
(c) What is the mean free path of a molecule in this container?
(d) When the block is remo
sure matches external air pressure. How much distance will the piston move during this
expansion?
(e) What is the work done on this gas by the surroundings while the piston moves to its new
position?
noved, the gas will expand isothermally until the internal pres-
Transcribed Image Text:(6) There is a cylindrical container with a piston with diameter d: volume to change. Sitting on top of that piston is a m = the diatomic gas is T 4.67cm which allows the 6.62kg block. The temperature of 382K and there are N 1.45 x 1023 molecules. || (a) What is the gauge pressure of the gas inside the container? (b) What is the volume of this container in cm³? (c) What is the mean free path of a molecule in this container? (d) When the block is remo sure matches external air pressure. How much distance will the piston move during this expansion? (e) What is the work done on this gas by the surroundings while the piston moves to its new position? noved, the gas will expand isothermally until the internal pres-
Expert Solution
Step 1

The block of mass of 6.62 kg will exert a force on the piston. In turn, the piston will exert a pressure on the gas in the cylinder. The pressure exerted on the piston due to the block is given as

P=FAF is the force exerted by the blockA is the area of the pistonP=mgπr2P=6.62×9.8π×(2.335×10-2)2P=64.8761.713×10-3P=3.7876×104 PaP=3.7876×104×9.86923×10-6 atmP=0.3738 atm

This is the gauge pressure inside the container

The absolute pressure of the gas in the container is

P=Pg+PaPa is the atmospheric pressureP=0.3738+1P=1.3738 atmP=1.37389.86923×10-6PaP=0.1392×106 Pa

This is the pressure of the gas inside the container

Step 2

The volume of the gas inside the container can be found using ideal gas law

PV=nRTn is he number of moles of the gasn can be calculated asn=Number of moleculesAvagadro's numbern=1.45×10236.022×1023n=0.24

PV=nRT0.1392×106 V=0.24×8.314×382V=762.227520.1392×106V=5475.772414×10-6 m3V=5475.772414 cm3

This is the volume of the gas present in the container

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