Relationship
between temperature, volume and pressure |
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Combining
Charles' and Boyle's Laws we get the expression on the right. In keeping
with the SI units V = volume in litres T = temperature in kelvin P = pressure in kPa. Although units are not critical we will stick to using kPa. |
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Changing
the temperature, pressure and volume of a fixed amount of gas the expression
on the left is true. Where P_{1} is the initial pressure and P_{2}
is the final pressure and so on. So let's put it to good use and see how it all works by trying some exercises below. |
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A fixed amount of gas was placed in a 6 litre vessel at 440 kPa at 30^{o}C . The vessel was left in the heat and allowed to expand to 10.5 litres at a pressure of 490 kPa. What is the temperature of the gas? |
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Step 1 Select the appropriate formula | ||
Step 2 Transform the formula to make it equal to temperature (T_{2}) | ||
Step
3 Replace the pronumerals on the right of the equation with their respective
values and solve for temperature. V^{1} = 6 L T^{1} = (30 + 273 ) = 303K P^{1} = 440 kPa V^{2} = 10.5 L P^{2} = 490 kPa T^{2} = ?^{o}K |
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1)
A fixed amount of gas is placed in a 2.5 litre vessel at 20^{o}C
and 95 kPa. The same gas is transferred into a 2 litre vessel at a temperature
of 40^{o}C . What is the pressure exerted by the gas? 2) 400 ml of pure hydrogen gas was collected at 27^{o}C and 90.5 kPa. It was later transferred into a 20 ml container at a temperature of -20^{o}C. What is the pressure of the gas? |
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3) One of the tyres of a car has a volume of 45 litres. The driver pumps the tyres up to a pressure of 320 kPa at 17^{o}C. After a long drive the temperature of the tyres increased to 27^{o}C . The tyre manufacturer recommends that the tyres not be inflated beyond 400 kPa as a blow out is likley to occur. Is the driver in danger of having a blow out? |