Combining the three laws of an ideal gas into one
Summarizing the gas laws dealt with, in our previous posts, we have PV = constant ( Boyle's law)
V / T = constant ( Charles's law)
P / T = constant ( Pressure law ).
These three equations can be combined into the single equation,
PV / T = constant
since, if T is constant, then PV = constent
or if P is constant, then V / T = constent
or if V is constant, then P / T = constant
It follows that P V / T = constant. includes all three gas laws.
since, if T is constant, then PV = constent
or if P is constant, then V / T = constent
or if V is constant, then P / T = constant
It follows that P V / T = constant. includes all three gas laws.
Now, with the solved problems
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| Solved problems |
This type of problems is solved by applying the relation
P / T = constant
P / T = constant
We, therefore write down the initial values and the required values of the volume, pressure and temperature, remembering that temperatures must be converted to the absolute ( Kelvin ) scale.
Initial values
P1 = 755 mmHg
V1 = 125 cm³
T1 = ( 273 + 15 ) = 288 K.
Required values
P2 = 760 mmHg
V2 = ?
T2 = ( 273 + 0 ) = 273 K.
Since P V/ T = constant , it follows that
P2 V2/ T2 = P1 V1 / T1
or V2 = V1 × P1 / P2 × T2 / T1
hence, substituting numerical values, we obtain
volume of gas at s.t.p. = V2 = 125 × 755 / 760 × 273/288
= 118 cm³.
P2 V2/ T2 = P1 V1 / T1
or V2 = V1 × P1 / P2 × T2 / T1
hence, substituting numerical values, we obtain
volume of gas at s.t.p. = V2 = 125 × 755 / 760 × 273/288
= 118 cm³.
B- An empty barometer tube, 1 m long, is lowered vertically, mouth downwards, into a tank of water. What will be the depth at the top of the tube when the water has risen 20 cm inside the tube? ( Atomspheric pressure may be assumed equal to 10.4 m head of water.)
Assuming the temperature remains constant, Boyle's law may be applied. i.e.,
Assuming the temperature remains constant, Boyle's law may be applied. i.e.,
P V = constant
or P1 V1 = P2 V2
Let h = depth, n m, of water-level in tube below surface, then
P2 = ( 10.4 + h ) in m of water
V2 = ( 0.8 × A ) in m³
where A = area of cross-section of tube in m² .
P1 = 10.4 in m of water
V1 = 1 × A in m³
Substituting in the above equation
( 10.4 + h ) ( 0.8 A ) = 10.4 × 1 × A
10.4 + h = 10.4 / 0.8 = 13
therefore h = 2.6 m
Hence top of tube is 2.6 - 0.8 = 1.8 m below surface.
Without using a mercury thermometer, we could calibrate the apparatus as a gas thermometer itself, simply by measuring the gas pressure first with the bulb in pure melting ice and then in steam. The Celsius scale of this thermometer is obtained by dividing the fundamental pressure interval into 100 equal parts, remembering that a suitable correction must be made if the steam is not at the standard pressure of 760 mmHg.
The most convenient way of representing the scale is to plot these two pressure values on a graph of pressure against temperature and to join them by a straight line.
or P1 V1 = P2 V2
Let h = depth, n m, of water-level in tube below surface, then
P2 = ( 10.4 + h ) in m of water
V2 = ( 0.8 × A ) in m³
where A = area of cross-section of tube in m² .
P1 = 10.4 in m of water
V1 = 1 × A in m³
Substituting in the above equation
( 10.4 + h ) ( 0.8 A ) = 10.4 × 1 × A
10.4 + h = 10.4 / 0.8 = 13
therefore h = 2.6 m
Hence top of tube is 2.6 - 0.8 = 1.8 m below surface.
C- When tested in a local garage at 10 °C a motor tyre is found to have a pressure of 1.2 N / cm² . Assuming the volume of the air inside remains constant, what would you expect the pressure to become after the tyre has been allowed to stand in the sun so that the temperature rises to 37 °C ? ( Atmospheric pressure = 1.0 N / cm² .
In this problem it must be understood that 1.2 N / cm² is the excess pressure above atmospheric pressure and hence the absolute pressure inside the tyre is ( 1.2 + 1.0 ) = 2.2 N / cm².
Applying the pressure law
P / T = constant
or P2 / T2 = P1 / T1
or P2 / T2 = P1 / T1
We have,
T2 = 273 + 37 = 310 K
P1 = 1.2 + 1.0 = 2.2 N / cm²
T2 = 273 + 37 = 310 K
P1 = 1.2 + 1.0 = 2.2 N / cm²
Substituting in the equation,
P2 / 310 = 2.2 / 283
whence P2 = 2.2 × 310 / 283 = 2.4 N / cm²
Therefore, new pressure as given by pressure gauge
= 2.4 - 1.0 = 1.4 N / cm².
P2 / 310 = 2.2 / 283
whence P2 = 2.2 × 310 / 283 = 2.4 N / cm²
Therefore, new pressure as given by pressure gauge
= 2.4 - 1.0 = 1.4 N / cm².
Jolly's constant volume apparatus as a thermometer in its own right
In the experiment with Jolly's apparatus to measure the presuure coefficient of expansion of air we used a mercury thermometer to measure temperature.Without using a mercury thermometer, we could calibrate the apparatus as a gas thermometer itself, simply by measuring the gas pressure first with the bulb in pure melting ice and then in steam. The Celsius scale of this thermometer is obtained by dividing the fundamental pressure interval into 100 equal parts, remembering that a suitable correction must be made if the steam is not at the standard pressure of 760 mmHg.
The most convenient way of representing the scale is to plot these two pressure values on a graph of pressure against temperature and to join them by a straight line.
When the bulb is at any other temperature we measure the pressure and ascertain the gas temperature by reference to the graph.
Now, if the gas in the tube were perfect then the result obtained would agree with the Kelvin scale which is independent of the properties of any particular substance.
Unfortunatly, no real gas is perfect but the one which comes most closely to this requirement is hydrogen.
An improved form of Jolly's apparatus is called the standard gas thermometer. This thermometer needs laborious corrections and is far too difficult and cumbersome for ordinary day-to-day use. When very high temperatures are to be measured it is filled with nitrogen and is employed only for the purpose of obtaining accurate values for a number of other fixed points both above 100 °C and below 0 °C, e.g., the freezing-point of gold ( 1063 °C ) and the boiling-point of oxygen ( - 193 °C ).
Unfortunatly, no real gas is perfect but the one which comes most closely to this requirement is hydrogen.
An improved form of Jolly's apparatus is called the standard gas thermometer. This thermometer needs laborious corrections and is far too difficult and cumbersome for ordinary day-to-day use. When very high temperatures are to be measured it is filled with nitrogen and is employed only for the purpose of obtaining accurate values for a number of other fixed points both above 100 °C and below 0 °C, e.g., the freezing-point of gold ( 1063 °C ) and the boiling-point of oxygen ( - 193 °C ).
In practice, therefore, such instruments as the platinum resistance thermometer, thermo-couples and so on are standardized by the fixed points so obtained and used to measure temperature over a wide range where mercury thermometers are unsuitable.
P.s. This article is available in pdf format. The page is named download pdf.
P.s. This article is available in pdf format. The page is named download pdf.