Question

If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12
liters, what is the temperature? ​

Answer 1

Explanation:

Let's model this gas as an "ideal gas". Then we can use the ideal gas law

PV=nRT

PV=nRT

In this case, here's what we know.

The amount of gas is n=4n=4 moles.

The pressure of gas is P=5.6P=5.6 atmospheres.

The volume of the gas is V=12V=12 liters.

The ideal gas constant always has a value of R=8.314R=8.314 joules per kelvin per mole. But we want that in terms of atmospheres and liters, so looking at the constant in different units on Wikipedia, we see that it is also 0.08210.0821 liter-atmospheres per kelvin per mole. Let's use that.

Now, let's solve our equation for the variable we want: TT.

T=PVnR

T=

nR

PV

And substitute in our values.

T=(5.6 atm)(12 L)(4 mol)(0.0821 L⋅atm⋅K−1⋅mol−1=5.6⋅124⋅0.0821 K=205 K

T

=

(4

mol

)(0.0821

L

⋅

atm

⋅K

−1

⋅

mol

−1

(5.6

atm

)(12

L

)

=

4⋅0.0821

5.6⋅12

K

=205 K

So the gas will have a temperature of around 205 kelvin which is -68.5 degrees Celsius.

Answer 2

The temperature is 210 K

Given - Number of moles, pressure, volume

Find - Temperature

Solution - The given variables are related to each other as follows -

PV = nRT

The value of R will be 0.08 L atm K-¹ mol-¹.

Keep the values in formula -

$T =\frac{PV}{nR}$

$T = \frac{5.6 \times 12}{4 \times 0.08}$

$T = \frac{67.2}{0.32}$

$T = 210 \: K$

The temperature is 210 K.

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