Question

At what temperature does 0.019135 moles of Ne in a 878.3 mL container exert a pressure of 0.946 atm?

Answer 1

Answer:

Temperature is 529.164 K

Explanation:

We are given

Number of moles of Ne (n) =  0.019135 moles

Volume (V) = 878.3 mL

Pressure (P) = 0.946 atm

We are required to calculate the temperature;

We can do this using the ideal gas law equation which is;

PV = nRT, where P is the pressure, n is the number of moles, V is the volume, R is the ideal gas constant (0.082057 Latm/mol/K) and T is the temperature.

From the equation;

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

$T = \frac{(0.946)(0.8783)}{(0.082057)(0.019135)}$

$T=529.164 K$

Therefore, the temperature will be 529.164 K.

Answer 2

Temperature of gas Ne : 529.208 K

Further explanation  

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated), among others  

• Boyle's law at constant T, P = 1 / V  
• Charles's law, at constant P, V = T  
• Avogadro's law, at constant P and T, V = n  

So that the three laws can be combined into a single gas equation, the ideal gas equation  

In general, the gas equation can be written  

$\large{\boxed{\bold{PV=nRT}}}$  

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08205 l.atm / mol K  

T = temperature, Kelvin  

Gas Ne with :

moles : 0.019135

volume : 878.3 ml = 0.8783 L

Pressure : 0.946 atm

Plug in gas ideal equation :

$\rm T=\dfrac{PV}{nR}\\\\T=\dfrac{0.946\times 0.8783}{0.019135\times 0.08205}\\\\T=\boxed{\bold{529.208\:K}}$

Learn more  

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