Question

What is the volume of .00831 mol of gas under ideal conditions if the

pressure is 1.01 atm and the temperature is 25 degrees C?*

Answer 1

Approximately
`0.20\; \rm L`.

Step-by-step explanation:

Convert the temperature of this gas to absolute temperature:

`T = 25\; \rm ^\circ C \approx (25 + 273.15)\; \rm K = 298.15\; \rm K`.

Let
`P` and
`V` represent the pressure and volume of this gas, respectively. Let
`n` represent the number of gas particles in this gas. Let
`R` represent the ideal gas constant. By the ideal gas law:

`P \cdot V = n \cdot R \cdot T`.

For this question:

• 
  `P = 1.01\; \rm atm` (given,)
• 
  `T = 298.15\; \rm K` (from unit conversion,) and
• 
  `n = 0.00831\; \rm mol`.

Look up the ideal gas constant
`R` that takes
`\rm atm` as the unit for pressure:

`R \approx 0.082057\; \rm L \cdot atm \cdot K^(-1)\cdot mol^(-1)`.

This question is asking for
`V`, the volume of this gas. Rearrange the ideal gas equation and solve for
`V`:

`\begin{aligned} V &= (n \cdot R \cdot T)/(P) \\ &\approx (0.00831\; \rm mol* 0.0082057\; \rm L \cdot atm \cdot K^(-1)\cdot mol^(-1)\cdot 298.15\; \rm K)/(1.01\; \rm atm) \\ &\approx 0.0020\; \rm L\end{aligned}`.