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14 g of nitrogen gas at STP are adiabatically compressed
Chapter 17, Problem 68P(choose chapter or problem)
Problem 68P
14 g of nitrogen gas at STP are adiabatically compressed to a pressure of 20 atm. What are (a) the final temperature, (b) the work done on the gas, (c) the heat input to the gas, and (d) the compression ratio V max /V min ? (e) Show the process on a pV diagram, using proper scales on both axes.
Questions & Answers
QUESTION:
Problem 68P
14 g of nitrogen gas at STP are adiabatically compressed to a pressure of 20 atm. What are (a) the final temperature, (b) the work done on the gas, (c) the heat input to the gas, and (d) the compression ratio V max /V min ? (e) Show the process on a pV diagram, using proper scales on both axes.
ANSWER:
Step 1 of 6
a.)
We have to find the final temperature of \(14 \mathrm{~g}\) of nitrogen gas at STP adiabatically compressed to a pressure of \(20 \mathrm{~atm}\).
The number of moles of nitrogen in \(14 \mathrm{~g}\) of nitrogen gas is
\(n=\frac{m}{M}\) Where, \(m=\) mass of nitrogen \(=14 \mathrm{~g}\) \(M=\) molecular mass of nitrogen \(=28 \mathrm{~g} / \mathrm{mol}\)
\(m=\) mass of nitrogen \(=14 \mathrm{~g}\)
\(M=\) molecular mass of
nitrogen \(=28 \mathrm{~g} / \mathrm{mol}\)
Thus,
\(n =\frac{14 \mathrm{~g}}{28 \mathrm{~g} / \mathrm{mol}}\)
= 0.5
The initial volume of nitrogen at STP that is \(T_{i}=273 \mathrm{~K}\) and \(p_{i}=1 \mathrm{~atm}=101300 \mathrm{~Pa}\) can be found using ideal gas equation.
\(p_{i} V_{i}=n R T_{i}\)
\(V_{i}=\frac{n R T_{i}}{p_{i}}\)
\(=\frac{(0.5)(8.31)(273)}{101300}\)
\(= 0.0112 \mathrm{~m}^{3}\)
\(V_{i} =11.2 \mathrm{~L}\)