Contents
Physics Topics are often described using mathematical equations, making them precise and quantifiable.
What is the Relation Between Expansion Coefficients?
From Charles’ law, if a fixed mass of gas at constant pressure occupies a volume V0 at 0°C, then at t °C its volume will be Vt = V0(1 + \(\frac{t}{273}\)).
Now if the temperature is reduced to t = -273°C, Vt = V0(1 – \(\frac{273}{273}\)) = 0, i.e., the volume of the gas vanishes at -273°C temperature.
Similarly from pressure law, if a fixed mass of a gas at constant volume has a pressure p0 at 0°C, then at t°C its pressure will be pt = p0(1 + \(\frac{t}{273}\)).
Now if the temperature is reduced to t = -273°C, pt = p0(1 – \(\frac{273}{273}\)) = 0, i.e., the pressure of the gas vanishes at -273°C temperature.
So, at -273°C, the volume and pressure of a fixed mass of gas has worked out to be zero. If the temperature could be lowered below -273°C, the pressure and volume would have been negative, which is meaningless. Hence, the lowest conceivable temperature in this universe is -273°C.
Therefore -273°C is called the absolute zero of temperature, or simply, absolute zero. More precisely, the value of absolute zero is -273.15C.
Definition: The temperature at which the volume and the pressure of a gas reduces to zero is called the absolute zero of temperature. This temperature is the lowest temperature in reality.
Taking absolute zero i.e., -273°C as zero, a new scale of temperature was developed by Lord Kelvin, and it is called the absolute scale of temperature or Kelvin scale after the inventor.
In this scale, the temperature reading is denoted by A (absolute) or K (Kelvin). Each degree in this scale is taken equal to a degree in Celsius scale. Hence, the scale of temperature where -273°C is taken as zero (0) and each degree is equal to a degree in Celsius scale, is called the absolute scale or Kelvin scale.
Freezing point of water (0°C) in absolute scale is 273 K and boiling point of water (100°C) in this scale is 373 K.
If any temperature is represented by t°C in Celsius scale and by TK in Kelvin scale, then, T = t + 273.
In Kelvin scale, a temperature may be zero or positive. A negative Kelvin temperature does not exist.
OK in Fahrenheit scale: In the relation \(\frac{C}{5}\) = \(\frac{F-32}{9}\), putting C = -273°C (0 K) we have, \(\frac{-273}{5}\) = \(\frac{F-32}{9}\) or, F= -459.4°F.
Incidentally, gas laws are valid as long as matter remains in a gaseous state. Actually, all gases get liquified much before they attain the absolute zero temperature. No gas can be cooled to the temperature of absolute zero, and as such, zero volume or zero pressure of a gas is never realised in practice.
Absolute scale of temperature—why it is so named: Selection of 0 degree in Celsius and Fahrenheit scale has no scientific reason behind it. Freezing point of water is taken arbitrarily as 0°C and boiling point as 100° C in Celsius scale. But selection of O degree in absolute scale is scientific as it denotes the temperature at which the volume and pressure of any ideal gas would reduce to zero. This is the lowest conceivable temperature in this universe. In addition, the scale does not depend on the nature of the gas. Hence, it is justified to call it the absolute scale of temperature.
Volume And Pressure Coefficients of a Gas
In general, both the volume and the pressure of a fixed mass of a gas undergo changes due to any change in its temperature. However, for convenience, we at first consider the two extreme types of heating :
1. either by keeping its pressure constant or
2. by keeping its volume constant. It is not possible to change the temperature of a gas keeping both pressure and volume constant. Hence, there are two coefficients of a gas. When a gas is heated, keeping the pressure constant, its volume increases and we get the volume coefficient at constant pressure (γp). Again when the gas is heated, keeping the volume constant, its pressure increases and we get the pressure coefficient at constant volume (γv).
Volume coefficient (γp) : The volume coefficient of a fixed mass of a gas at a constant pressure is the increment of its volume when the temperature of a unit volume is raised by 1°C from 0°C.
Let at constant pressure, volume of a specific amount of gas be V0 at 0°C and Vt at t °C.
∴ Increase in volume = Vt – V0 and increase in temperature = t – 0 = t°C
∴ Increase in volume for 1°C rise in temperature of a unit volume of the gas = \(\frac{V_t-V_0}{V_0 t}\)
i.e., the volume coefficient γp = \(\frac{V_t-V_0}{V_0 t}\) …… (1)
or, Vt = V0(1 + γpt) ……. (2)
Pressure coefficient (γv): The pressure coefficient of a fixed mass of a gas at a constant volume, initially at unit pressure, is the increment of its pressure when its temperature is raised by 1°C from 0°C.
Let at a constant volume the pressure of a specific amount of gas be p0 at 0°C and pt at t°C.
∴ Increase in pressure = pt – p0
and increase in temperature = r – 0 = t °C
∴ Increase in pressure for 1°C rise in temperature per unit initial pressure of the gas = \(\frac{p_t-p_0}{p_0 t}\)
i.e., pressure coefficient, γv = \(\frac{p_t-p_0}{p_0 t}\) … (3)
or, pt = p0(1 + γvt) …….. (4)
Relationship between the two coefficients of expansion of a gas: Suppose at 0°C a fixed mass of gas has volume V0 and pressure p0. The gas is heated to t °C. We can perform this increase in temperature in any of the two ways [Fig.].
i) Keeping the volume V0 constant when pressure increases to pt from p0.
ii) Keeping the pressure p0 constant when volume increases to Vt from V0
Following pressure law in the first case, pt = p0(1 + γvt)
In the second case as per Charles’ law,
Vt = V0(1 + γpt)
Since the final temperature is t°C in both the cases, as per Boyle’s law
ptV0 = p0Vt or, p0V0(1 + γvt) = p0V0(1 + γpt)
∴ γv = γp
Hence, for any Ideal gas the coefficient of volume expansion is equal to the coefficient of pressure expansion.
A comparison of equations (2) and (4) respectively with the corresponding expressions derived from Charles’ law (Section 6.3, equation 1) and pressure law (Section 6.4, equation 1) shows that
γp = \(\frac{1}{273}\) or, 0.00366 °C-1
and also γv = \(\frac{1}{273}\) or, 0.00366 °C-1
Therefore,
1. the pressure coefficient (γv) and the volume coefficient (γp) have the same value which is same for all gases, though different solids and liquids have different values for the coefficients of volume expansion.
2. Gases are heated in a container like liquids. But the volume expansion of a gas is much higher (near about 100 times) than the corresponding expansion of the container. Unless much accuracy is required, two separate expansion coefficients (real and apparent) are not needed. Practically, apparent expansion coefficient of a gas is the same as real expansion coefficient of the gas.
3. To find the volume coefficient (γp) of a gas, initial volume is to be taken at 0°C and
4. to find the pressure coefficient of a gas, initial pressure is to be taken at 0°C.
To illustrate 3 or 4 the following example may be considered:
Let the initial volume V0 of a fixed mass of a gas at 0°C, be 273 cm3.
According to Charles’ law,
volume at 100°C, V100 = v0(1 + \(\frac{100}{273}\))
= 273(1 + \(\frac{100}{273}\)) = 373 cm3
and volume at 150°C, V150 = V0(1 + \(\frac{150}{273}\))
= 273(1 + \(\frac{150}{273}\)) = 423cm3
In case of solids and liquids magnitudes of expansion coefficients are too small. So to find volume expansion in case of solids and liquids it is not always necessary to take initial volume at 0°C. Values of expansions do not differ much if we consider initial volume at some temperature other than 0°C.
Numerical Examples
Example 1.
Volume of a gas Is doubled by raising its temperature at constant pressure. Initial temperature of the gas was 13°C. Find the final temperature.
Solution:
As the pressure is constant, using Charles’ law we have, \(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\)
Here V1 = x cm3 (suppose), T1 = 273 + 13 = 286 K and V2 = 2x cm3.
∴ \(\frac{x}{286}\) = \(\frac{2 x}{T_2}\) or, T2 = 572 K = (527 – 273)°C = 299°C.
Example 2.
The volume of fixed mass of a gas at 47°C is 640cm3 and its pressure is 75 cm of Hg. To which temperature should the gas be raised at constant volume to make its pressure double?
Solution:
As volume is a constant, we get, using pressure law
\(\frac{p_1}{T_1}\) = \(\frac{p_2}{T_2}\)
Here p1 = 75 cmHg, T1 = 273 + 47 = 320K
and p2 = 2 × 75 = 150 cmHg.
\(\frac{75}{320}\) = \(\frac{150}{T_2}\)
or, T2 = 640 K = (640 – 273)°C = 367°C.
Example 3.
The volume of a fixed mass of gas is 300 cm3 at STP. When the temperature is raised to 50°C at constant volume, the pressure exerted by the gas becomes 900 mmHg. What is the pressure coefficient of the gas?
Solution:
Here, pt = 900 mmHg, po = 760 mmHg and t = 50°C.
∴ pt = p0(1 + γvt)
∴ γv = \(\frac{p_t-p_0}{p_0 t}\)
or γv = \(\frac{900-760}{760 \times 50}\) = \(\frac{140}{760 \times 50}\) = 0.00368°C-1.
Example 4.
At constant pressure, 1f the volume of a fixed mass of gas at temperature 80°C is 500 cm3 and that at 150°C is 600 cm3, what is the coefficient of volume expansion (γp) of the gas?
Solution:
We have, Vt = V0(1 + γpt)
Using the given conditions we get,
500 = V0(1 + γp × 80) …….. (1)
and 600 = V0(1 + γp × 150) …… (2)
Dividing (2) by (1), we get,
\(\frac{6}{5}\) = \(\frac{1+150 \gamma_p}{1+80 \gamma_p}\) or, 6 + 480γp = 5 + 750γp
or, 270γp = 1 or, γp = \(\frac{1}{270}\)°C-1
Example 5.
If heated to 35°C at constant pressure, the volume of gas increases from 5 L at 0°C, 640 cm3. What should be the value of absolute zero for this gas in Celsius scale?
Solution:
Let the absolute zero temperature for that gas be -T°C.
So, 0°C = TK = T1, 35°C = (T + 35) K = T2,
V1 = 5000 cm3 and V2 = 5000 + 640 = 5640 cm3.
As per Charles’ law, \(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) at constant pressure.
∴ \(\frac{5000}{T}\) = \(\frac{5640}{T+35}\) or, 500(T + 35) = 564T
64 T = 17500, T = \(\frac{17500}{64}\) = 273.43
Hence, absolute zero in Celsius scale = -273.43°C.
Example 6.
A hydrogen cylinder can withstand an internal pressure of 7 × 106 Pa. The pressure of hydrogen in cylinder at 15°C is 1.7 × 106 Pa. At what minimum temperature an explosion may take place?
Solution:
Given, p1 = 1.7 × 106 Pa
and T1 = 273 + 15 = 288 K
Explosion may occur at a pressure p2 = 7 × 106 Pa
As volume is constant in the cylinder, from pressure law,
\(\frac{p_1}{T_1}\) = \(\frac{p_2}{T_2}\) or, T2 = \(\frac{p_2 T_1}{p_1}\)
∴ T2 = \(\frac{7 \times 10^6 \times 288}{1.7 \times 10^6}\) = 1185.9K
= (1185.9 – 273)°C = 927°C
Example 7.
A glass vessel is filled with air at 30 °C. Up to which temperature should the vessel be heated keeping the pressure constant so that \(\frac{1}{3}\) rd of the initial volume of air is expelled? γp = \(\frac{1}{273}\)°C-1. [HS ‘02]
Solution:
Let initial volume of air = V1
∴ Final volume of an equal mass of air,
V2 = V1 + \(\frac{V_1}{3}\) = \(\frac{4}{3}\)V1
[as volume of expelled air = \(\frac{1}{3}\)V1]
Initial temperature, T1 = 273 + 30 = 303 K. Let the required temperature be T2 K.
As pressure is constant,
\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) or, \(\frac{V_1}{303}\) = \(\frac{\frac{4}{3} V_1}{T_2}\)
or, T2 = \(\frac{4}{3}\) × 303 = 404K = (404 – 273)°C = 131°C.
Example 8.
At 27°C, and at a pressure of 76 cmHg 100 cm3 of a gas is collected over water surface. The space occupied by the gas is saturated with water vapour. Maximum vapour pressure of water at 27°C is 17.4 mmHg. What will be the volume of dry gas at STP?
Solution:
Let the volume of the dry gas at STP = V2 cm3, pressure p2 = 76 cmHg and temperature T2 = 0°C = 273 K.Given V1 = 100 cm3, p1 = 76 – 1.74 = 74.26 cmHg and T1 = 27 + 273 = 300K.
Hence, from the equation of state,
\(\frac{p_1 V_1}{T_1}\) = \(\frac{p_2 V_2}{T_2}\) or, \(\frac{100 \times 74.26}{300}\) = \(\frac{V_2 \times 76}{273}\)
or, V2 = \(\frac{273 \times 74.26}{3 \times 76}\) = 88.92 cm3.
∴ At STP, the volume of dry gas will be 88.92 cm3.
Example 9.
A person measures the pressure of his car tyre to be 2 × 105 Pa. At that time the temperature and pressure of the atmosphere are 27°C and 1 × 105 Pa respectively. Then he travels to another city where the temperature and pressure of the atmosphere are 12°C and 6.7 × 1o4 Pa respectively. Then what will be the pressure of his car tyre at that time. Assume the volume of the tyre is same in both cases.
Solution:
The pressure in a tyre is a guage pressure, which is the difference between the pressure in the tyre and atmospheric pressure.
Hence, absolute pressure in the tyre = guage pressure + atmospheric pressure.
So in 1st case,
absolute pressure, p1 = 2 × 105 + 105 = 3 × 105 Pa
and temperature, T1 = 273 + 27 = 300 K
Let in the 2nd case the measured pressure (guage pressure) = x Pa.
So the absolute pressure, p2 = (x + 6.7 × 104) Pa
and temperature, T2 = 273 + 12 = 285 K
Since the volume of the tyre is constant, \(\frac{p_1}{T_1}\) = \(\frac{p_2}{T_2}\)
or, \(\frac{3 \times 10^5}{300}\) = \(\frac{x+6.7 \times 10^4}{285}\) or, x = 2.18 × 105
So the measured pressure is 2.18 × 105 Pa.