Question

nitrogen gas at standard atmospheric pressure 101.3kPa has a volume of 0.080m^3 . if there are 3.0 mol of gas what is the temperature of nitrogen gas

Answer 1

Approximately
`52\; {\rm ^\circ C}` (approximately
`325\; \rm K`), assuming that nitrogen is an ideal gas.

Step-by-step explanation:

• Let
  `P` denote the pressure of this nitrogen gas sample.
• Let
  `V` denote the volume of this nitrogen gas sample.
• Let
  `n` denote the number of moles of
  `\rm N_(2)` molecules in this nitrogen gas sample.
• Let
  `T` denote the absolute temperature of this nitrogen gas sample (typically measured in degrees kelvins.)

Let
`R` denote the ideal gas constant. By the ideal gas law, the following equation would relate these quantities:

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

Rearrange this equation to obtain an expression for
`T`:

`\begin{aligned}T &= (P \cdot V)/(n \cdot R)\end{aligned}`.

Look up the ideal gas constant:
`R \approx 8.314\; \rm Pa \cdot m^(3) \cdot K^(-1) \cdot mol^(-1)`.

Convert each measurements from the question to standard units:

• 
  `P = 101.3\; \rm kPa = 101.3 * 10^(3)\; \rm Pa`.
• 
  `V = 0.080\; \rm m^(3)`.
• 
  `n = 3.0\; \rm mol`.

Substitute these values into the expression for
`T`:

`\begin{aligned}T &= (P \cdot V)/(n \cdot R) \\ &\approx (101.3* 10^(5)\; \rm Pa * 0.080\; \rm m^(3))/(3.0\; \rm mol * 8.314\; \rm Pa \cdot m^(3) \cdot K^(-1) \cdot mol^(-1)) \\ &\approx 324.91\; \rm K\end{aligned}`.

Convert the unit of this temperature to degrees celsius:

`\begin{aligned} & 324.91\; \rm K \\ =\; & (324.91 - 273.15)\; {\rm ^\circ C} \\ \approx \; & 52\; {\rm ^\circ C} \end{aligned}`.