Percentage change in the density of a gas

In summary, the conversation discusses finding the percentage change in density when moving a gas from one vessel to another at the same temperature, and whether increasing the temperature in the second vessel will affect the gas's density. The summary also mentions using Boyle's law to determine the change in density, and corrects some mistakes in the initial calculations.
  • #1
patric44
296
39
Homework Statement
a gas was at a specific vessel at T=300K, P=110KPa, when moved to another different vessel at the same temperature its pressure becomes 105KPa, if this vessel was heated to 330K find the percentage change in the density?
Relevant Equations
P1V1=P2V2, V1/T1=V2/T2
Hi all, in this question i was asked to find the percentage change in the density, my approach was as following, first i find the change in volume due to putting the gas into the other vessel as:
$$
P_{1}V_{1}=P_{2}V_{2}\;\; → \;\;V_{2}=\frac{P_{1}}{P_{2}}V_{1}
$$
now i use
$$
V_{1}/T_{1}=V_{2}/T_{2}\;\; → \;\;\frac{P_{1}}{P_{2}}V_{1}/T_{1}=V_{2}/T_{2}
$$
and using V=m/rho leads to, with m constant, i arrive at
$$
\frac{\rho_{o}-\rho_{2}}{\rho_{o}}=\frac{1}{\frac{T_{2}P_{1}}{T_{1}P_{2}}}-1=-13.22%
$$
the answer has the choices [-5.55% , -4.55% , 4%, 5%]
what I am doing wrong, can any one help
 
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  • #2
Hi. The question says the gas is moved to "another different vessel at the same temperature". This is unclear.

It sounds like the temperature immediately after the move is still 300K. So the gas is heated from 300K to 330K after the move.

Will increasing the temperature of the gas in the 2nd vessel affect the gas's density?
 
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  • #3
Steve4Physics said:
Hi. The question says the gas is moved to "another different vessel at the same temperature". This is unclear.

It sounds like the temperature immediately after the move is still 300K. So the gas is heated from 300K to 330K after the move.

Will increasing the temperature of the gas in the 2nd vessel affect the gas's density?
yes it was 300K then heated to 330K, yes I guess, since the increase in temperature will make the gas expand and hence lower its density
 
  • #4
patric44 said:
yes it was 300K then heated to 330K, yes I guess, since the increase in temperature will make the gas expand and hence lower its density
The vessel itself doesn't expand (or its expansion is negligible).

Will increasing the temperature change the gas's volume?
 
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  • #5
wait a minute, the other container has a fixed volume hence the gas has no place to expand when heated! what was I thinking :woot:, so the change in density only comes from the movment in the other vessel which I can determine by Boyle's law, thanks so much its clear now
 
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  • #6
patric44 said:
$$\frac{\rho_{o}-\rho_{2}}{\rho_{o}}=..$$
By the way, a 'change' is [final value]- [initial value]. So the change in density is ##\rho_{final} - \rho_{initial}##, not the other way round.

Also, the symbol for 'kilo' is lower case 'k' (e.g. 110kPa).

(For information, where a unit is named after someone, the full unit name is given in lower case and the (first letter of) the symbol is given in upper case. E.g. K, kelvin; Hz, hertz; and you can amuse yourself by thinking of others!)
 
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