Finding both temperature and the amount of gas added

[lorenz0]

Homework Statement

$30g$ of $O_2$ are inside a cylinder with a radius of $0.1 m$ and a height of $0.7 m$ at a temperature of $20^{\circ}C$. Calculate the gas pressure. Oxygen is added until the pressure increases by $80\%$ compared to the initial one. Calculate the final temperature and the amount of oxygen introduced into the cylinder.

Relevant Equations

$PV=nRT$

The volume of the cylinder is $V=\pi r^2 h=\frac{7\pi}{250}\ m^3$ the number of moles is $n=\frac{15}{16}\ mol$ so from $PV=nRT$ we get $P=\frac{nRT}{V}=25975.5\ Pa$.

Now, for the second question, it should be an isochoric process so $V_2=V_1$ and $P_2=P_1+0.8P_1=\frac{9}{5}P_1$ and from these, using the ideal gas law, I get $n_1T_1=\frac{5}{9}n_2P_2$ but I have two variables ($n_2$ and $T_2$) so I don't see how I could solve this and I would thus appreciate an hint about how to go forward, thanks.

 

[Chestermiller]

Homework Statement:: $30g$ of $O_2$ are inside a cylinder with a radius of $0.1 m$ and a height of $0.7 m$ at a temperature of $20^{\circ}C$. Calculate the gas pressure. Oxygen is added until the pressure increases by $80\%$ compared to the initial one. Calculate the final temperature and the amount of oxygen introduced into the cylinder.
Relevant Equations:: $PV=nRT$

The volume of the cylinder is $V=\pi r^2 h=\frac{7\pi}{250}\ m^3$ the number of moles is $n=\frac{15}{16}\ mol$ so from $PV=nRT$ we get $P=\frac{nRT}{V}=25975.5\ Pa$.

Now, for the second question, it should be an isochoric process so $V_2=V_1$ and $P_2=P_1+0.8P_1=\frac{9}{5}P_1$ and from these, using the ideal gas law, I get $n_1T_1=\frac{5}{9}n_2P_2$ but I have two variables ($n_2$ and $T_2$) so I don't see how I could solve this and I would thus appreciate an hint about how to go forward, thanks.

There’s not enough information given to answer this. If you make some assumptions you can use the open system version of the first law thermodynamics to answer this.

 

[haruspex]

You seem to have used a radius of 0.2m.
For the second part, you are not told the temperature and pressure of the incoming gas, nor whether it is isothermal, adiabatic, or between the two.
I guess you'll have to assume it comes in at the current pressure and temperature at each stage, and since it asks about change in temperature we can dismiss isothermal.
If that's all correct, you can't solve it just by plugging in the standard adiabatic equations. You'll need to consider adding a small amount and integrate up.

Edit: I see Chet has replied. I defer to Chet.

 

[Steve4Physics]

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Homework Statement:: $30g$ of $O_2$ are inside a cylinder with a radius of $0.1 m$ and a height of $0.7 m$ at a temperature of $20^{\circ}C$. Calculate the gas pressure. Oxygen is added until the pressure increases by $80\%$ compared to the initial one. Calculate the final temperature and the amount of oxygen introduced into the cylinder.
Relevant Equations:: $PV=nRT$

The volume of the cylinder is $V=\pi r^2 h=\frac{7\pi}{250}\ m^3$ the number of moles is $n=\frac{15}{16}\ mol$ so from $PV=nRT$ we get $P=\frac{nRT}{V}=25975.5\ Pa$.

Just to add that "25975.5" has six significant figures. You may want to reflect on this.

 

[lorenz0]

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Just to add that "25975.5" has six significant figures. You may want to reflect on this.

Thanks; I knew but I was so worried about not being able to understand what was happening in the second part of the problem that I didn't take the time to put the right number of significant figures in my initial calculations. I will obviously correct this when I write up the problem in my notes.

 

[MRMMRM]

0.02199 m^3 Volume of cylinder (it should be 7*pi/1000.)

293.00K =20C

30g of 32g O2=0.9375 mol

8.3144598 J/mol K = R

103,864 pascal = initial pressure

I only added this as I think you got the volume wrong. +80%

186,955 pascal

PV/R=nT

PV/R= (186,955*.02199)/8.3144598= 494.457 mol K

From here you would need to be given or assume the final temperature or the final moles of O2.

some examples:

494.457 mol*K/ 0.9375mol = 527.4K, This is no additional oxygen just an increase in temperature.

494.457 mol*K/ 293K= 1.69mol, 1.69mol-.9375mol= 0.75mol, this is just additional O2 with no temperature change.

They only stated an increase in oxygen, so you could assume the temperature even decreased, just as long as the oxygen remained a gas and did not become a liquid near 55*K to 73*K(-218*C to -200*c) at this given pressure (I'm not exactly sure what the condensing point for oxygen at 1.9 bar is, it could be much higher). You may need to add ~6 to 8 moles near these temperatures to get the 80% desired increase in pressure.

So how much oxygen was added for an 80% increase in pressure could be from 0 to .75 moles, all the way up to 6-8 moles with a temperature decrease depending on when it started to turn into a liquid.

 

[Chestermiller]

As I said in my previous response, you can answer this question if you make some reasonable assumptions about the oxygen being added, by means of the open system (control volume) version of the 1st law of thermodynamics. From the open system version of the first law, we have that $$d(nu)=h_{in}dn\tag{1}$$where u is the internal energy per mole of the oxygen in the cylinder, n is the number of moles of oxygen in the cylinder, and $h_{in}$ is the enthalpy per mole of oxygen injected into the cylinder. This assumes that the cylinder is adiabatic.

For an ideal gas, we can write $$u=C_v(T-T_{ref})\tag{2}$$and $$h_{in}=C_v(T_{in}-T_{ref})+Pv=C_v(T_{in}-T_{ref})+RT_{in}\tag{3}$$where $T_{ref}$ is an arbitrary reference temperature for zero internal energy and $T_(in)$ is the temperature of the injected oxygen. Substitution of Eqns. 2 and 3 into Eqn. 1 yields: $$nC_vdT=[C_v(T_{in}-T)+RT_{in}]dn$$If we now assume that the injected oxygen is at the same room temperature value (293 K) as the original oxygen in the cylinder, then we have:$$nC_vdT=(293C_p-C_vT)dn$$or, equivalently, $$\frac{dT}{(293\gamma-T}=\frac{dn}{n}$$Integrating this between the initial and final state yields: $$\frac{293\gamma-T_1}{293\gamma-T_2}=\frac{n_2}{n_1}$$with $T_1=293$. So we have $$(\gamma-1)T_1n_1=(\gamma T_1-T_2)n_2$$This, together with $\frac{T_2n_2}{T_1n_1}=1.8$ enables us to solve for $T_2$ and $n_2$: $$\frac{T_2}{T_1}=\frac{1.8\gamma}{1.8+\gamma-1}$$