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 the temperature? Express your answer to two significant figures

Answer 1

Answer:

$\large \boxed{\text{490 ^{\circ} C}}$

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data  

p = 28 atm

V = 0.045 L

n = 0.020 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

$\begin{array} {rcl}pV & = & nRT\\\text{28 atm} \times \text{0.045 L} & = & \rm\text{0.020 mol} \times 0.08206 \text{L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\1.26&=&0.001 641T\text{ K}^{-1}\\T& = &\dfrac{1.26}{\text{0.001 641 K}^{-1}}\\\\\end{array}$

$\begin{array} {rcl} & = & \text{767.7 K}\\\end{array}\\T = \text{(767.7 - 273.15) ^{\circ} C} = \large \boxed{\textbf{490 ^{\circ} C}}$