a gas has a volume of 1.75 L at -23 degrees Celsius and 150 kPa at what temperature will the gas occupy 1.30 L at 210 kPa
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To determine the temperature at which the gas will occupy 1.30 L at 210 kPa, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (150 kPa)
V1 = initial volume (1.75 L)
T1 = initial temperature (-23 degrees Celsius)
P2 = final pressure (210 kPa)
V2 = final volume (1.30 L)
T2 = final temperature
Let's plug in the given values into the equation and solve for T2:
(150 kPa * 1.75 L) / (-23°C) = (210 kPa * 1.30 L) / T2
To simplify the equation, let's convert temperatures from Celsius to Kelvin:
T1 = -23°C + 273.15 = 250.15 K
Now, rearranging the equation to solve for T2:
(150 kPa * 1.75 L * T2) = (210 kPa * 1.30 L * 250.15 K)
Simplifying further:
T2 = (210 kPa * 1.30 L * 250.15 K) / (150 kPa * 1.75 L)
T2 ≈ 351.216 K
Therefore, the gas will occupy a volume of 1.30 L at 210 kPa at a temperature of approximately 351.216 K.