## Q. 12.12

Calculating the Temperature of a Gas Using the Ideal Gas Law

Nitrous oxide, N$_2$O, is naturally present, in trace amounts, in the atmosphere. Its source is decomposition reactions occurring in soils. What would be the temperature, in degrees Celsius, of a 67.4 g sample of N$_2$O gas under a pressure of 5.00 atm in a 7.00 L container?

## Verified Solution

The amount of N$_2$O present is given in grams rather than moles. The grams need to be changed to moles prior to using the ideal gas law.

$67.4 \cancel{\textrm{g N}_2 \textrm{O}} \times \frac{1 \textrm{mole N}_2\textrm{O}}{44.02 \cancel{\textrm{g N}_2 \textrm{O}}} = 1.5311222$ moles N$_2$O      (calculator answer)
= 1.53 moles N$_2$O          (correct answer)

Three of the four variables in the ideal gas equation (P, V, and n) are now known, and the fourth (T) is to be calculated.

P = 5.00 atm                   n = 1.53 moles
V = 7.00 L                      T = ? K

Rearranging the ideal gas equation to isolate T on the left side gives

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

Since the pressure is given in atmospheres and the volume in liters, the value of R to be used is

$R = 0.08206 \frac{\textrm{atm} \cdot \textrm{L}}{\textrm{mole} \cdot \textrm{K}}$

Substituting numerical values into the equation gives

$T = \frac{(5.00 \cancel{\textrm{atm}}) (7.00 \cancel{\textrm{L}})}{(1.53 \cancel{\textrm{moles}}) \Bigl(0.08206 \frac{ \cancel{\textrm{atm}} \cdot \cancel{\textrm{L}}}{\cancel{\textrm{mole}} \cdot \textrm{K}} \Bigr) }$

Notice again how the gas constant units, except for K, cancel. After cancellation, the expression 1/(1/K) remains. This expression is equivalent to K. That this is the case can be easily shown. All we need to do is multiply both the numerator and denominator of the fraction by K.

$\frac{1 \times \textrm{K}}{\frac{1}{\cancel{\textrm{K}}} \times \cancel{\textrm{K}}} = K$

Doing the arithmetic, we get as an answer 279 K for the temperature of the N$_2$O gas.

$T = \frac{(5.00) (7.00)}{(1.53) (0.08206)}$ K = 278.7694 K      (calculator answer)