Question

What would be the temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg

Answer 1

The temperature would be 922.01 °K

Step-by-step explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law:

P*V=n*R*T

Where P is the gas pressure, V is the volume it occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

• P= 2300 mmHg
• V= 15 L
• n= 0.6 moles

• R= 62.36367
  `(mmHg*L)/(mol*K)`

• T=?

Replacing:

2300 mmHg* 15 L=0.6 moles*62.36367
`(mmHg*L)/(mol*K)` *T

Solving:

`T=(2300 mmHg*15 L)/(0.6 moles*62.36367 (mmHg*L)/(mol*K))`

T= 922.01 °K

The temperature would be 922.01 °K