Answer: The temperature in degrees Celsius is 51.21ºC.
Step-by-step explanation:
An ideal gas is a hypothetical gas whose pressure-volume-temperature
behavior can be completely accounted for by the ideal gas equation:
![p×V = n×R×T](https://img.qammunity.org/2020/formulas/chemistry/high-school/e7drlb8e441x4dv3e0z6dofc3802kh4j5e.png)
We want to know the temperature of the sample, which will be in Kelvin, so we rearrange the equation to give:
![T= p×V ÷ n×R](https://img.qammunity.org/2020/formulas/chemistry/high-school/t3basg4om3xyr20l6kncs6mwziajidq16m.png)
Knowing that:
p= 2.69 atm
n= 6.50 mol
V= 64. 27 L
and R, the gas constant is 0.082 L× atm ÷ mol×K
![T = 2.69 atm×64.27 L ÷ 6.50 mol×0.082 L.atm÷mol.K = <strong>324.36 K</strong>](https://img.qammunity.org/2020/formulas/chemistry/high-school/6p0v8o9aic1ccg03wbq25okl704riu0fqp.png)
Now, we need to convert Kelvins to degrees Celsius. We use the following equation:
![K = (ºC + 273.15 ºC) × 1K ÷ 1ºC](https://img.qammunity.org/2020/formulas/chemistry/high-school/gz5umvzlz72je28flt5voso7gzlurv5fkg.png)
K= 324.36 K - 273.15ºC = 51.21 ºC
Therefore, the temperature of the sample is 51.21 degrees Celsius.