Question

Find the volume of a gas STP if it’s volume is 80.0 mL at 109 kpa and-12.5c

Answer 1

Approximately
`8.38 * 10^8\; \rm mL`, which is the same as
`8.38 * 10^5 \; \rm L`. Assumption: the behavior of this gas is ideal.

Step-by-step explanation:

Initial state of the gas:

• 
  `T_\text{initial} = -12.5\; \rm ^\circ C = (-12.5 + 273.15)\; \rm K = 260.65\; \rm K`.
• 
  `P_\text{initial} = \; \rm 10^9\; \rm kPa = 10^(12)\; \rm Pa`.

STP state:

• 
  `T_\text{STP} = 0\; \rm ^\circ C = 273.15\; \rm K`.
• 
  `P_\text{STP} = 10^5\; \rm Pa`.

The initial state of this gas can be changed to STP state in two steps:

• First, reduce the pressure from
  `10^(12)\; \rm Pa` to
  `10^5\; \rm Pa`.
• Second, increase the temperature from
  `260.65\; \rm K` to
  `273.15\; \rm K`.

Assume that the gas acts like an ideal gas at all time. Also, assume that the number of gas particles did not change.

Assume that temperature stays the same when the pressure changes from
`10^(12)\; \rm Pa` to
`10^5\; \rm Pa`. By Boyle's Law, volume is inversely proportional to pressure when all other factors stay the same. In other words,

`\begin{aligned}V_\text{intermediate} &= V_\text{initial} \cdot \displaystyle \frac{P_{\text{initial}}}{P_{\text{STP}}} \\ &= 80.0\; \rm mL * (10^(12)\; \rm Pa)/(10^5\; \rm Pa) = 8.00 * 10^8\; \rm mL\end{aligned}`.

After that, assume that pressure stays the same when the temperature changes from
`\rm 260.65\; \rm K` to
`\rm 273.15\; \rm K`. By Charles's Law, volume is proportional to pressure when all other factors stay the same. In other words,

`\begin{aligned}V_\text{STP} &= V_\text{intermediate} \cdot \displaystyle \frac{T_{\text{STP}}}{T_{\text{initial}}} \\ &= 8.00 * 10^8\; \rm mL * (273.15\; \rm K)/(260.65\; \rm K) \\ &\approx 8.38* 10^8\; \rm mL = 8.38 * 10^5\; \rm L\end{aligned}`.