Question

The temperature and pressure at Shimla are 15 degree Celsius and 72.0 cm of mercury and at Kalka these are 35 degree Celsius and 76 cm of mercury. Find the ratio of air density at Kalka to the air density at Shimla.

Answer 1

Hint: In this question, we will use the relation between pressure, volume, temperature, mass and gas constant. This will give us an expression relating the air density. Further, we can use the resultant expression to find the ratios of the air densities of Shimla and Kalka respectively.

Formula used:
$PV = \left( {\dfrac{m}{M}} \right)RT$

Complete step by step solution:
As we know that the temperature is different for different regions. Temperature is a physical quantity, and it determines the hotness and coolness. S.I unit of temperature is Kelvin, which is represented as K.
Here, we have the relation between pressure, volume, temperature, mass and gas constant, which is given as:
$\eqalign{& PV = \left( {\dfrac{m}{M}} \right)RT \cr
  & \Rightarrow P = \left( {\dfrac{m}{V}} \right)\left( {\dfrac{1}{M}} \right)RT \cr} $
Now we take the ratio of the two different air densities:
$\eqalign{& \Rightarrow P = \left( {\dfrac{d}{M}} \right)RT \cr
  & \Rightarrow d = \dfrac{{PM}}{{RT}} \cr} $
$\left( {\dfrac{{{d_1}}}{{{d_2}}}} \right) = \left( {\dfrac{{{P_2}}}{{{T_2}}}} \right) \times \left( {\dfrac{{{T_1}}}{{{P_2}}}} \right)$
$ \Rightarrow \left( {\dfrac{{{d_1}}}{{{d_2}}}} \right) = \left( {\dfrac{{76 \times 288}}{{308 \times 72}}} \right)$
$\therefore \left( {\dfrac{{{d_1}}}{{{d_2}}}} \right) = 0.987$
Therefore, we get the ratio between the two different air densities of Shimla and Kalka in the above equation.

Additional Information: Pressure is defined as the force exerted on an object per unit area. Unit of pressure is Pascal which is represented as P. The unit of pressure can also be written as Newton per meter square.
We know that there are four laws of thermodynamics, the first law of thermodynamics states that internal energy change of a system equals net heat transfer minus net work done by the system.
There are four processes of heat transfer i.e., isobaric process, isothermal process, isochoric and adiabatic process.
We know that, in an isochoric process the volume is held constant, meaning that the work done by the system will be zero. This process is also known as an isometric process or an iso- volumetric process.
An isothermal process is a change of a system, in which the temperature remains constant- ΔT = 0. This happens when a system is in contact with an outside thermal reservoir or a heat bath, and the change occurs slowly enough to allow the system to continually adjust to the temperature of the reservoir through heat exchange. We know that an adiabatic process is where a system exchanges no heat with its surroundings.

Note: The four laws of thermodynamics are first law, second law, third law and the 0th law. The heat in adiabatic processes is zero. Also, the subtraction of specific heat capacity at constant pressure and specific heat capacity at constant volume gives the gas constant.