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At what temperature (in K) would 2.10 mol of N2 gas have a pressure of 1.25 atm in a 22.1 L tank? (R = 0.0821 L·atm/mol·K.

Question

At what temperature (in K) would 2.10 mol of N2 gas have a pressure of 1.25 atm in a 22.1 L tank? (R = 0.0821 L·atm/mol·K.

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Solution

To solve this problem, we can use the ideal gas law equation, which is:

PV = nRT

Where: P = pressure V = volume n = number of moles R = ideal gas constant T = temperature

We are asked to find the temperature (T), so we rearrange the equation to solve for T:

T = PV / nR

Now we can substitute the given values into the equation:

T = (1.25 atm * 22.1 L) / (2.10 mol * 0.0821 L·atm/mol·K)

Now, calculate the value:

T = 27.75 / 0.17241

T = 160.9 K

So, the temperature of the gas would be approximately 161 K.

This problem has been solved

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