# Question 90f5f

May 8, 2017

The final pressure will be $\text{7.40 atm}$.

#### Explanation:

The Combined Gas Law can be used to answer this question. It states that the volume of a given amount of gas is proportional to the ratio of its Kelvin temperature and its pressure. The equation is:

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

At STP, the standard temperature is ${0}^{\circ} \text{C}$ or $\text{273.15 K}$. With gas laws, the Kelvin temperature is used. The standard pressure can vary, but since you need the answer in atmospheres, it will be $\text{1 atm}$. (The more current standard pressure is $\text{10"^5color(white)(.) "Pa}$, or $\text{100 kPa}$, or $\text{1 bar}$, all of which are equal.

To convert $\text{^@"C}$ to Kelvins, add $273.15$ to the Celsius temperature.

Given
${P}_{1} = \text{1 atm}$
${V}_{1} = \text{0.500 L}$
${T}_{1} = \text{273.15 K}$
${V}_{2} = 7.50 \times {10}^{- 2} \textcolor{w h i t e}{.} \text{L"="0.0750 L}$
${T}_{2} = \text{30"^@"C"+273.15="303 K}$

Unknown: ${P}_{2}$

Solution
Rearrange the equation to isolate ${P}_{2}$ and insert the given data into the equation, then solve.

${P}_{2} = \frac{{P}_{1} {V}_{1} {T}_{2}}{{T}_{1} {V}_{2}}$

P_2=(1"atm"xx0.500color(red)cancel(color(black)("L"))xx303color(red)cancel(color(black)("K")))/(273.15color(red)cancel(color(black)("K"))xx0.0750color(red)cancel(color(black)("L")))="7.40 atm"# rounded to three significant figures

May 8, 2017

$\text{739.7 kPa}$

#### Explanation:

The gas law states that $P V = n R T$.

In the first part of the question, you are told that the gas occupies $\text{0.500 L}$ at standard temperature and pressure.
this means that conditions are:

• $\text{273.15 Kelvin}$ (0 celsius)
• $\text{100,000 Pascals}$ of pressure
• $\text{0.500 L}$ volume (Which corresponds to $\text{0.0005 m"^3}$)
The $R$ in the gas law is the Molar gas constant, this value is $\text{8.314 m"^3 "Pa K"^(−1) "mol"^(−1)}$ .

With this information, you can calculate the number of moles of gas you have in your sample.

$P V = n R T$

$n = \frac{P V}{R} T$

$n = \frac{100 , 000 \cdot 0.0005}{273.15 \cdot 8.314}$

$\text{n=0.02201 Moles}$

With this information, you can now answer the second step. The conditions are now:

• $\text{303.15 Kelvin}$
• $\text{0.075 L}$ (which corresponds to $\text{0.000075 m"^3}$)
-$\text{0.02201 Moles}$
Now the equation is re-arranged to $P = \frac{n R T}{V}$

$P = \frac{0.02201 \cdot 8.314 \cdot 303.15}{0.000075}$

$P = \text{739650.19 Pa}$

The final pressure is $\text{739650 Pascals}$ or $\text{739.7 kPa}$. This is $\text{7.30 atm}$.

Hope this helped :D