Question

a 75.0 - liter container contains 15.82 moles of argon at a pressure of 546.8 kilopascals. what is the temperature of the container? type the correct answer in the box. express your answer to three significant figures. the temperature of the container is k. celsius to kelvin conversion: k = °c + 273.15

Explanation:

Step1: Identify the formula

Use the ideal - gas law $PV = nRT$.

Step2: Convert units

Convert pressure to Pa ($546.8\ kPa = 546800\ Pa$) and volume to $m^{3}$ ($75.0\ L=0.075\ m^{3}$).

Step3: Substitute values

Substitute $P = 546800\ Pa$, $V = 0.075\ m^{3}$, $n = 15.82\ mol$, and $R = 8.314\ Pa\cdot m^{3}/(mol\cdot K)$ into $T=\frac{PV}{nR}$.

Step4: Calculate

Calculate the value of $T$ and round to three significant figures.

Answer:

We first use the ideal - gas law $PV = nRT$ to find the temperature in Celsius and then convert it to Kelvin.

1. Rearrange the ideal - gas law $PV=nRT$ for $T$. The ideal - gas constant $R = 8.314\ J/(mol\cdot K)=8.314\ Pa\cdot m^{3}/(mol\cdot K)$. First, convert the pressure $P = 546.8\ kPa=546800\ Pa$ and the volume $V = 75.0\ L = 0.075\ m^{3}$, and $n = 15.82\ mol$.

• From $PV=nRT$, we have $T=\frac{PV}{nR}$.
• Substitute the values: $T=\frac{546800\ Pa\times0.075\ m^{3}}{15.82\ mol\times8.314\ Pa\cdot m^{3}/(mol\cdot K)}$.
• Calculate the numerator: $546800\times0.075 = 41010$.
• Calculate the denominator: $15.82\times8.314=131.52748$.
• Then $T=\frac{41010}{131.52748}\ K\approx312\ K$.
• We can also use the conversion formula $K = ^{\circ}C+273.15$. If we want to double - check using the Celsius scale first:
• From $T=\frac{PV}{nR}$, after getting $T$ in Kelvin, we can convert it to Celsius. First, assume we calculate $T$ in Celsius using $T_{C}=\frac{PV}{nR}-273.15$. But since the problem doesn't ask for Celsius and we can directly get the answer in Kelvin from the ideal - gas law calculation.

1. Rounding to three significant figures, the temperature of the container is $312\ K$.