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A sample of gas is put into a rigid (fixed volume) container at -3 oC and a pressure of 37.3 kPa. The container is then placed in an oven at 267 oC.

What pressure would you expect to measure for the gas in the container at this higher temperature?

kPa

1 Answer

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Step-by-step explanation:

The ideal gas law can be used to calculate the pressure of a gas at a given temperature and volume. The ideal gas law is given by the equation PV = nRT, where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas.

To solve this problem, we need to first convert the temperature in degrees Celsius to degrees Kelvin. This is because the ideal gas constant has units of energy per mole per Kelvin, and the temperature in the ideal gas law must be in Kelvin. To convert from degrees Celsius to degrees Kelvin, we simply add 273.15 to the temperature in degrees Celsius. Therefore, the temperature of the gas in Kelvin is -3 + 273.15 = 270.15 K.

Next, we need to know the volume of the gas in the container. If the container is rigid, then the volume of the gas will not change, regardless of the temperature of the gas. Therefore, we can assume that the volume of the gas is the same at both temperatures. We can also assume that the number of moles of gas in the container remains constant, since the container is sealed and no gas can escape or enter the container.

With this information, we can plug the values into the ideal gas law equation and solve for the pressure at the higher temperature:

PV = nRT

P = (nRT) / V

P = (n * 8.314 * 270.15) / V

P = (n * 8.314 * 267) / V

We know the value of V, the volume of the gas, and the value of n, the number of moles of gas, but these values are not given in the problem. Therefore, we cannot solve for the pressure of the gas at the higher temperature without knowing these values.

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