Thermodynamics/Gas Law questions

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SUMMARY

This discussion focuses on solving two thermodynamics problems related to ideal gases. For the first question, the user correctly applied the ideal gas law (n = PV/RT) to determine the number of moles of gas as 22.22. To find the mass, they need to multiply the number of moles by the molecular weight of 352 g/mol, resulting in a mass of 7777.84 grams. The density can then be calculated using the formula density = mass/volume. In the second question, the user attempted to find the volume change using the equation P1/T1 = P2/T2, leading to the conclusion that the volumes remain approximately the same.

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Carbonoid
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Hi,

I'm stuck on two questions, I gratefull if someone could help me with them, firstly;

(Q1) "A cubical container of volume 0.10m^3 contains a gas at 300K and 1000000Pa. Assuming that the gas is ideal, determine
a) the number of moles of gas present, given that the Universal Gas constant is 8.3 J/K/mol.
b) the mass of gas present given that its relative molecular mass is 352.
c) the density of the gas"

I'm okay I think working out 1a I used the equation n=PV/RT and got 22.22. I can't seem to find how to work out the mass...and 1c I think I'm okay, I guess I just use M/V.

Secondly;

(Q2) "A fixed mass of gas in passing through a jet engine has its pressure increased from 300000 to 1300000Pa, while its temperature rises from 80(deg C) to 1500 (deg C). By what factor does the volume of gas change?"

I worked out an answer to this but it looks wrong, I used P1/T1 = P2/T2 and found the difference between those two to be 116.638, so is that the correct method to work out the change of volume?

Thanks again
 
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(1) Hint: What is the mass of one mole of a material with molecular weight 352?

(2)[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex] so [tex]\frac{3 \times 10^5 V_1}{353}=\frac{1.3 \times 10^6 V_2}{1773}[/tex] so [tex]V_1=\frac{353 \times 1.3 \times 10^6} {1773 \times 3 \times 10^5} V_2 \approx \frac{5 \times 10^8}{5 \times 10^8} V_2 = V_2[/tex]
so I get that they should be roughly the same.
 
hey thanks, I get it now
 

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