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2 years ago

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A sample of oxygen gas (O2) has a volume of 7.51 L at a temperature of 19°C and a pressure of 1.38 atm. Calculate the moles of O

2 molecules present in this gas sample.

1 answer:

Maslowich2 years ago

8 0

Answer: 0.43molO₂

Explanation:

The ideal gas law for moles is $n=\frac{PV}{RT}$ . If you did not know, R is the ideal gas constant, $0.0821\frac{L*atm}{mol*K}$ .

Since the temperature must be in Kelvin, we must convert our given temperature from ℃ to K. $19C+273=292K$

Now that we have the temperature converted to Kelvin, we can plug in our information to the ideal gas law for the number of moles.

$n=\frac{PV}{RT}$

$n=\frac{1.38atm*7.51L}{0.0821\frac{L*atm}{mol*K} *292K} =0.43molO_{2}$

$n=0.43molO_{2}$

Therefore, the number of moles is 0.43molO₂.

I hope this helps! Pls mark brainliest!! :)

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A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

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Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

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