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Abd.alhourani

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According to the Ideal Gas Law (from Physics and Chemistry), the pressure of a body of gas (P) with a volume (V) and temperature (T) mu following equation:
PV=KT
for a known gas constant k. Pis measured in atmosphere a (Atm), Vis measured in liters and Tis measured in degrees Celsius C)
Suppose a 10-liter tank is filled with a gas which is at 1 atmosphere and has a temperature of 25°C. If the temperature of the gas in heated up at a rate of 5°C/min. answer the following questions: (a) At what rate is the pressure changing when the temperature is 45°C


a) At what rate is the pressure changing when the temperature is 45 cn Answer:
b) Does the rate of change for the pressure depend on the temperature? Support your answer.
 
According to the Ideal Gas Law (from Physics and Chemistry), the pressure of a body of gas (P) with a volume (V) and temperature (T) mu following equation:
PV=KT
for a known gas constant k. Pis measured in atmosphere a (Atm), Vis measured in liters and Tis measured in degrees Celsius C)
Suppose a 10-liter tank is filled with a gas which is at 1 atmosphere and has a temperature of 25°C. If the temperature of the gas in heated up at a rate of 5°C/min. answer the following questions: (a) At what rate is the pressure changing when the temperature is 45°C


a) At what rate is the pressure changing when the temperature is 45 cn Answer:
b) Does the rate of change for the pressure depend on the temperature? Support your answer.
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Be careful! You have "\(\displaystyle \frac{dt}{dt}\)" which looks like it would be 1. But you mean "\(\displaystyle \frac{dT}{dt}\)", where T is the temperature and t is the time. It is important to make it clear that those are different variables that represent different amounts.

You have PV= KT where P is the pressure, V the volume, T the temperature, and K is a constant. Further, because the gas is in a 10 liter tank, the volume is constant also. Differentiating both sides with respect to time, t, \(\displaystyle \frac{d(PV)}{dt}= V\frac{dP}{dt}= K\frac{dV}{dt}\).

You are told that "a 10-liter tank is filled with a gas which is at 1 atmosphere and has a temperature of 25°C" so with T= 25, V= 10, and P= 1. We have PV= 10= KT= 24K. K= 10/24= 5/12.

When the temperature is 45° C, volume is still 10 and K is still 5/12 so we have
\(\displaystyle 10\frac{dP}{dt}= \frac{5}{12}\frac{dT}{dt}\).

The pressure is increasing at 5° C/min so dT/dt= 5.

Solve \(\displaystyle 10\frac{dP}{dt}= =\frac{5}{12}(5)= \frac{25}{12}\) for dP/dt.
 
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