A gas at 61°C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L?
constant pressure...
V1/V2=T1/T2
.67/1.12=(61+273)/(273+Tc)
solve for Tc , the celcius temp for the new volume.
/ is divide?
To solve this problem, we can use the combined gas law, which relates the initial and final temperatures and volumes of a gas. The combined gas law equation is as follows:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
Where:
P₁ and P₂ are the initial and final pressures of the gas (which can be assumed to be constant),
V₁ and V₂ are the initial and final volumes of the gas,
T₁ and T₂ are the initial and final temperatures of the gas.
In this case, we are given:
V₁ = 0.67 L
V₂ = 1.12 L
T₁ = 61°C (convert to Kelvin for calculations)
To convert Celsius to Kelvin, we will use the formula:
T(Kelvin) = T(Celsius) + 273.15
So, T₁ = 61°C + 273.15 = 334.15 K
Now we can rearrange the combined gas law equation to solve for T₂:
T₂ = (P₂ * V₂ * T₁) / (P₁ * V₁)
Since the pressure is assumed to be constant, we can omit it from the equation:
T₂ = (V₂ * T₁) / V₁
Plugging in the values we have:
T₂ = (1.12 L * 334.15 K) / 0.67 L
Now we can perform the calculations:
T₂ ≈ 558.29 K
Finally, to convert the temperature back to Celsius, we subtract 273.15:
T₂ ≈ 558.29 K - 273.15 ≈ 285.14°C
Therefore, at approximately 285.14°C, the volume will increase to 1.12 L.