A compressor takes 0.50 m3 of a gas at 33°C and 760 mm Hg and compresses it to 0.10 m3, cooling it to -55°C at the same time. What is the pressure of the gas at these new conditions?
Question
A compressor takes 0.50 m3 of a gas at 33°C and 760 mm Hg and compresses it to 0.10 m3, cooling it to -55°C at the same time. What is the pressure of the gas at these new conditions?
Solution
To find the pressure of the gas at the new conditions, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where: P1 = initial pressure of the gas (760 mm Hg) V1 = initial volume of the gas (0.50 m3) T1 = initial temperature of the gas (33°C + 273.15 = 306.15 K) P2 = final pressure of the gas (unknown) V2 = final volume of the gas (0.10 m3) T2 = final temperature of the gas (-55°C + 273.15 = 218.15 K)
Now, let's substitute the given values into the equation:
(760 mm Hg * 0.50 m3) / (306.15 K) = (P2 * 0.10 m3) / (218.15 K)
Simplifying the equation:
380 mm Hg = (P2 * 0.10 m3) / (218.15 K)
To find P2, we can rearrange the equation:
P2 = (380 mm Hg * 0.10 m3) / (218.15 K)
Calculating the value:
P2 = 174.32 mm Hg
Therefore, the pressure of the gas at the new conditions is approximately 174.32 mm Hg.
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