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For many purposes we can treat propane (C3H8) as an ideal gas at temperatures above its boiling point of −42.°C. Suppose the temperature of a sample of propane gas is raised from −10.0°C to 3.0°C, and at the same time the pressure is increased by 10.0%.

User Robodisco
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Final answer:

The temperature and pressure increase in propane is studied using the ideal gas law. As temperature rises, so does pressure, and this is further compounded by an additional increase in pressure.Thus, in this case, the pressure of the propane would increase not only because of the temperature increase but also because of the additional 10% pressure increase.

Step-by-step explanation:

The question relates to the behavior of a gas, specifically propane (C3H8), under certain conditions of temperature and pressure. In this case, the temperature of the propane is raised from −10.0°C to 3.0°C and at the same time its pressure is increased by 10.0%.

According to the ideal gas law, with the formula PV=nRT, where P represents pressure, V volume, n is the number of moles, R is the ideal gas constant, and T is temperature. If the amount of gas and the volume stays constant, as temperature goes up, the pressure goes up, and vice versa. Thus, in this case, the pressure of the propane would increase not only because of the temperature increase but also because of the additional 10% pressure increase.

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User Crazyscot
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