Answer:
494.1 kPa
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (kPa)
P2 = final pressure (kPa)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question,
P1 = 294 kPa
P2 = ?
V1 = 42.9 liters
V2 = 22.8 liters
T1 = 76.0°C = 76 + 273 = 349K
T2 = 38.7°C = 38.7 + 273 = 311.7K
294 × 42.9/349 = P2 × 22.8/311.7
12612.6/349 = 22.8 P2/311.7
36.14 = 22.8P2/311.7
Cross multiply
36.14 × 311.7 = 22.8P2
11264.605 = 22.8P2
P2 = 11264.605 ÷ 22.8
P2 = 494.1 kPa
Answer:
Q = 143,921 J = 143.9 kJ.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:

Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:

Regards!
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