Answer :
Final answer:
The temperature of 100.8 g of N2 gas that exerts a pressure of 23.8 atm in a 6.51 L container is approximately 1396.91 K.
Explanation:
To find the temperature of the N2 gas, we can use the ideal gas law equation pV = nRT, where p is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in kelvin. In this case, the pressure is 23.8 atm, the volume is 6.51L, and the molar mass of N2 is 28.01 g/mol. First, we need to calculate the number of moles of N2 using the given mass.
$\text{moles} = \frac{\text{mass}}{\text{molar mass}}$
$\text{moles} = \frac{100.8\,\text{g}}{28.01\,\text{g/mol}}$
$\text{moles} \approx 3.6\,\text{mol}$
Now we can plug the values into the ideal gas law equation and solve for the temperature.
$pV = nRT$
$23.8\,\text{atm} \times 6.51\,\text{L} = 3.6\,\text{mol} \times R \times T$
From the given information, we know that the gas constant R is 0.0821 L·atm/(mol·K). We can rearrange the equation to solve for temperature:
$T = \frac{{pV}}{{nR}}$
$T = \frac{{23.8\,\text{atm} \times 6.51\,\text{L}}}{{3.6\,\text{mol} \times 0.0821\, \text{L·atm/(mol·K)}}}$
$T \approx 1396.91\,\text{K}$
Therefore, the temperature of 100.8 g of N2 gas that exerts a pressure of 23.8 atm in a 6.51 L container is approximately 1396.91 K.
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