## Chapter 12

## 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?

## Step-by-Step

## 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

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

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}}} = KDoing 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)

**= 279 K (correct answer)**

The calculated temperature is in kelvins. To convert to degrees Celsius, the unit specified in the problem statement, we subtract 273 from the Kelvin temperature.

** T(°C)** = 279 K – 273 =

**6°C**

**Answer Double Check**: Is the magnitude of the numerical answer reasonable? Yes. Rounding off the input numbers for the calculation to numbers easier to work with—5, 10, 2, and 0.1—the ballpark answer for the problem is (5 × 10)/(2 × 0.1) = 250. The value of the calculated answer, in kelvins, is of the same order as the rounded answer.