Exactly one mole of an ideal gas is contained in a 2.00 liter container at 1,000 K. What is the pressure exerted by this gas?

2 answers:

lisabon 2012 [21]3 years ago

6 0

PV = nRT
P = ???
V = 2 liters.
n = 1 mol
T = 1000 K
R = 0.08205

P = nRT/V
P = 1 * 0.08205 * 1000 / 2
P = 41.03 atmospheres. That's a pretty big pressure. <<<< answer.

Kipish [7]3 years ago

4 0

pV = nRT

p = nRT/V

p= 1 x 0.08205 x 1000/ 2

p = 41.025 Pa

Edit: The unit should be atm instead of Pa, as pointed out by a nice human being.

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Calculate the molar volume occupied by 1 mole of N2 using the van der Waals equation in the form of virial expansion at (a) its

dimulka [17.4K]

Answer:

Explanation:

Given that:

Pressure P = 10 atm

number of moles of N $_2$ = 1 mole

correction for the attractive force between molecule a = 1.352

correction for the volume of molecules b = 0.0387 L mol-1

Rate = 8.314

critical Temperature Tc= 126.3 K

Using van deer Waal equation:

$(p+ \dfrac{n^2a}{V^2} ) (V-nb)=nRT$

So for 1 mole of N $_2$ gas; we have

$(p+ \dfrac{a}{V^2} ) (V-b)=RT$

$(10+ \dfrac{1.352}{V^2} ) (V-0.0387)=8.314 \times 126.3$

Recall that: Tc = $\dfrac{ 8a}{27Rb}$

and

$Pc=\dfrac{a}{27b^2}$

Vc = 3b

Thus;

Vc= 3×(0.0387)

Vc = 0.1161m³

using the above Van der Waal equation, the value of V can also be determined, but in several situations, we abandon the term of (V - nb) and sometimes abandon $(p+n^2*\dfrac{a}{V^2})$ depending on the circumstances.