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A gas occupies 400.0 mL at 20.00 C and 70.00 kPa. What will the pressure be at 40.00 C and 200.0 mL ?

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Answer: The new pressure is 280 kPa.

Step-by-step explanation:

Given:
V_(1) = 400.0 mL,
T_(1) = 20^(o)C ,
P_(1) = 70.0 kPa


V_(2) = 200.0 mL,
T_(2) = 40.0^(o)C,
P_(2) = ?

Now, combined gas law is used to calculate the new pressure as follows.


(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))

Substitute the values into above formula as follows.


(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))\\(70.0 kPa * 400.0 mL)/(20^(o)C) = (P_(2) * 200.0 mL)/(40.0^(o)C)\\P_(2) = 280 kPa

Thus, we can conclude that the new pressure is 280 kPa.

User Dviljoen
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