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A 3.950 l sample of gas is cooled from 91.50°c to a temperature at which its volume is 2.550 l. what is this new temperature? assume no change in pressure of the gas.

User TonE
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Final answer:

The new temperature of the gas after its volume is reduced from 3.950 L to 2.550 L is approximately -36.64°C, according to Charles's Law.

Step-by-step explanation:

The question concerns the relationship between gas volume and temperature, assuming no change in pressure. The scenario outlined falls under Charles's Law, which states that the volume of a gas is directly proportional to its temperature on the Kelvin scale. The equation for Charles's Law is V1/T1 = V2/T2, where V1 and T1 represent the initial volume and temperature, and V2 and T2 represent the final volume and temperature.

To find the new temperature (T2), we must first convert the initial temperature from Celsius to Kelvin by adding 273.15: 91.50°C + 273.15 = 364.65 K. Using the initial volume (3.950 L) and the final volume (2.550 L), we can plug these into the equation: (3.950 L/364.65 K) = (2.550 L/T2). Solving for T2 gives us T2 = (2.550 L × 364.65 K) / 3.950 L. Calculating T2, we get T2 ≈ 236.51 K, which is approximately -36.64°C.

User Bodo
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