Question

veredAnswerQuestion 3Suppose the gas inside a closed, steel cylinder has a pressure of 6.1 atm at2.0 °C. What will be the pressure (atm) of the gas at -83.0 °C? Enter youranswer to 2 decimal places.-710/1 pts4.22 margin of error +/- 0.05

Answer 1

4.22 atm

Step-by-step explanation

Given that:

The initial pressure, P₁ = 6.1 atm

The initial temperature, T₁ = 2.0 °C = (2.0 + 273.15 K) = 275.15 K

The final temperature, T₂ = -83.0 °C = (-83.0 + 273.15 K) = 190.15 K

What to find:

The final pressure, P₂ (atm) of the gas at 83.0 °C.

Step-by-step solution:

The relationship between pressure and temperature of gaseous subtance was decribed by Amonton's gas law.

Amonton's Law states that the pressure of an ideal gas varies directly with the absolute temperature when the volume of the sample is held constant.

Mathematically, we have:

`(P_1)/(T_1)=(P_2)/(T_2)`

Putting the values of the parameters into the formula, we have:

`\begin{gathered} \frac{6.1\text{ }atm}{275.15\text{ }K}=\frac{P_2}{190.15\text{ }K} \\ \\ P_2=(6.1atm*190.15K)/(275.15K) \\ \\ P_2=4.22\text{ }atm \end{gathered}`

Thus, the final pressure, (atm) of the gas at 83.0 °C is 4.22 atm.