Question

A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pressure is constant at 3 atm. What was the original temperature of the gas?

Answer 1

Answer: The original temperature was

$T_{1}=126.51K$

Explanation:

Let's put the information in mathematical form:

$V_{1}=12.7m^{3}$

$T_{1}=?$

$V_{2}=32.5m^{3}$

$T_{2}=323K$

$P_{1}=P_{2}=3atm$

If we consider the helium as an ideal gas, we can use the Ideal Gas Law:

$PV=nRT$

were R is the gas constant. And n is the number of moles (which we don't know yet)

From this, taking $R=0.08205746\frac{atm.l}{mol.K}$,  we have:

$n=\frac{P_{2}V_{2}}{RT_{2}}$

⇒$n=3.67mol$

Now:

$T_{1}=\frac{P_{1}V_{1}}{nR}$

⇒$T_{1}=126.51K$

Answer 2

Answer:

126.22

Explanation:

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