Question

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

Answer 1

Answer:

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

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,

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

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,

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