Help me out with two thermodynamic questions?

Question

1)
A 20-cm-diameter cylinder that is 40 cm long contains 50 g of oxygen gas at 20°C.

a) What is the number density of the oxygen?
___________________ m^-3

b) What is the reading of a pressure gauge attached to the tank?
_______________________kPa


2)
A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm ) to the surface (where the pressure is 1.00 atm ). The temperature at the bottom is 4.0°C , and the temperature at the surface is 23.0°C.
What is the ratio of the volume of the bubble as it reaches the surface (Vs) to its volume at the bottom (Vb)?
Vs/Vb = ___________________________


thanks

Answer 1

1)

a)

Volume of the cylinder

= π r^2 h

= π (10)^2 * 40 cc

= 4π liter

50 g of oxygen = 50/32 g-mol = 5/32 x 6.022 x 10^23 molecules



=> number of molecules / m^3

= (50/32 x 6.022 x 10^23) / (4π x 10^-3)

= 7.4877 x 10^25 molecules / m^3.



b)

Volume of the cylinder

= π r^2 h

= π (10)^2 * 40 cc

= 4π liter



50g = 50/32 g-mol



As a chemical engineer, I am accustomed to use a thumb rule that 1 g-mole of an ideal gas occupies 22.4 liter at 0° C and 1 atm (101.325 kPa)



=> (50/32) g-mol will occupy

(50/32) * 22.4 * (273 + 20)/(273) liter at 20° C and 1 atm

= 37.5641 liter



As the actual volume is 4π liter, its pressure will be

= (37.5641/4π) x 101.325 kPa

= 302.9 kPa. absolute

= 302.9 - 101.325 kPa gauge pressure

= 201.6 kPa gauge pressure.



Edit:

2)

Vs/Vb

= (273+23)/(273+4) * * (3.50/1.0)

= 3.74.



Edit:

Let me recheck part 1b) by another method

For an ideal gas,

PV = nRT

=> 4π P = (50/32) * (0.082) * (273 + 20)

=> P = (50/32) * (0.0821) * (273 + 20) / 4π atm

=> P = 2.991

=> P = 2.991 * 101.325 kPa = 303.1 kPa. absolute pressure

=> P = 303.1 - 101.3 kPa

=> P = 201.8 kPa.



Edit:

What I had found is the absolute pressure. You have asked gauge pressure which is 1 atm = 101 kPa less

=> answer is 202 kPa.

Answer 2

I'm sorry but I don't get question 1a.



For 1b,

The volume of the cylinder is 0.01256 <-- I got this from V = pi*r^2*h



Where the molar mass of oxygen is 16g/mol

Therefore, the number of moles of oxygen in this case is 3.125 mols



The temperature of the gas is 293K <--- 20C + 273K = 293K



Therefore, knowing the equation PV = nRT

P = (nRT)/V

Substituting the above values will yield you your answer



For the second question, use the formula Ts/Tb = Vs/Vb, but note that the temperature has to be in kelvin. This can be done by adding 273K to each of the temperature values.



Good luck!

Answer 3

i'm not sure in this one. i'm in basic terms acquainted with the logarithm appearing in isothermal approaches, no longer isobaric ones. because of the fact the stress is continuous right here, the exchange in volume could be calculated without postpone. V = nRT/P V1 = (2 mol) R (300K) / (1500 J/m^3) V1 = 3.326 m^3 V2 = (2 mol) R (400K) / (1500 J/m^3) V2 = 4.434 m^3 dV = (V2 - V1) dV = (4.434m^3 - 3.326m^3) dV = one million.108 m^3 So i could then artwork it as: Q = (3/2) (2 mol) R (100K) + (1500J/m^3) (one million.108m^3) Q = 2494J + 1662J Q = 4156J I hate thinking there's a malicious program interior the answer (plenty greater possibly my attempt is incorrect), yet i think it rather is achievable that somebody substituted an isothermal equation quite the isobaric one? EDIT: Ah, confident. From the different comments it rather is extremely not an appropriate gasoline, so my calculations are not valid. So under no circumstances recommendations!

Answer 4

We can use the dimensions of the cylinder given to us to determine the volume which is .05 m^3



We know that one mol of oxygen has a mass of 32 grams



32/1 = 50/x ; x=1.56 mols



so there's 1.56 mols of oxygen in the .05m^3 tank



1.56/.05 = x/1 ; x = 31.2 mols m^-3



If you want molecules m^-3 just multiply 31.2 by Avogadro's number.



For part b we can use PV = nRT , we know everything except pressure so we can solve for it. Just make sure units are consistent, i.e. temperature must be in Celsius. If we're measuring the gauge pressure then take that pressure and subtract atmospheric pressure.



For part 2:



(Pb)(Vb) = nR(Tb)

(Ps)(Vs) = nR(Ts)



divide them:



((Pb)(Vb))/((Ps)(Vs)) = (Tb)/(Ts)



Vb/Vs = ((Tb)(Ps))/((Ts)(Pb))



Vs/Vb = ((Ts)(Pb))/((Tb)(Ps))