Question

The gas in an engine cylinder is compressed until it has a volume of 0.045 liter and an absolute pressure of 28 atmospheres. If the engine cylinder contains 0.020 mole of gas, what is its temperature?

Answer 1

$7.7 \times 10^{2} \; \text{K}$.

Explanation

Units of each quantity:

• P = 28 atm;
• V = 0.045 L;
• n = 0.020 mol;

Ideal gas constant:

$R = 0.0820573 \;\text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$.

Apply the ideal gas law:

$T = \dfrac{P \cdot V}{n \cdot R} = \dfrac{{\bf 28} \; \text{atm} \times {\bf 0.045} \;\text{L} }{{\bf 0.020}\; \text{mol} \times {\bf 0.0820573} \;\text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}} = {\bf 767.76} \; \text{K}$.

All data given in this question come with two significant figures. Round the value of T to two significant figures:

$T = 767.76 \;\text{K} = 7.7 \times 10^{2} \; \text{K}$.