Question

Find the dimensions of :- a) specific heat capacity C, (b) The coefficient of linear expansion $$\alpha$$, (c) the gas constant R. Some of the equations involving these quantities are $$Q=mc(T_2-T_1)$$, $$I_t=I_o[1+\alpha (T_2-T_1)]$$, $$PV=nRT$$

Answer 1

Hint: In this question we have been asked to find the dimensions of the specific heat capacity, the coefficient of linear expansion and the gas constant R. we have been given three different equations containing these parameters individually. Therefore, we shall use the equations and the dimensions of other parameters for dimensional analysis and thus calculate the dimension of the said parameters.

Complete step by step solution:
a) We have been asked to calculate the specific heat capacity C. the equation given to us is,
$$Q=mc(T_2-T_1)$$
We know that, in above equation
Q is the amount of heat energy measured in Joules
m is the mass of system measured in kg
$$(T_2-T_1)$$ is the temperature difference in kelvin (dimension of temperature is given as $$\theta$$)
Now, solving for C
We get,
$$c={\displaystyle \frac{Q}{m(T_2-T_1)}}$$
Writing the above equation in form of unit
We get,
$$c={\displaystyle \frac{joules}{kg\times \theta }}$$
We know that,
$$Joules=Nm$$
Therefore, dimensions of specific heat will be given as,
$$c={\displaystyle \frac{\left[M^1{L}^1{T}^{-2}\right]\left[M^0{L}^1{T}^0\right]}{\left[M^1{L}^0{T}^0\right]\left[M^0{L}^0{T}^0{\theta }^1\right]}}$$
Therefore,
On solving,
$$c=\left[M^0{L}^2{T}^{-2}{\theta }^{-1}\right]$$ ………………… (1)
b) here we have been asked to calculate the dimensions of coefficient of linear expansion
we have been given equation,
$$l_t=l_o[1+\alpha (T_2-T_1)]$$
Where, l is the length and T is the temperature having dimension as $$\theta$$
Now, solving above equation for $$\alpha$$
We get,
$$\alpha ={\displaystyle \frac{l_t-l_o}{l_o(T_2-T_1)}}$$
Therefore, the dimensions of $$\alpha$$ are given as,
$$\alpha ={\displaystyle \frac{\left[M^0{L}^1{T}^0\right]}{\left[M^0{L}^1{T}^0\right]\left[M^0{L}^0{T}^0{\theta }^1\right]}}$$
Therefore,
$$\alpha =\left[M^0{L}^0{T}^0{\theta }^{-1}\right]$$ ………………….. (2)
c) here we have been asked to find the dimensions of the universal gas constant R from the equation
$$PV=nRT$$
Solving for R,
$$R={\displaystyle \frac{pV}{nT}}$$
Pressure is given as ratio of force and area, in dimensional form
We get,
$$p={\displaystyle \frac{\left[ML{T}^{-2}\right]}{L^2}}$$
Therefore,
$$p=\left[M^1{L}^{-1}{T}^{-2}\right]$$
Also,
$$V=L^3$$ and
$$n=\left[mol\right]$$
After substituting above values in ideal gas equation for R
We get,
$$R={\displaystyle \frac{\left[M^1{L}^{-1}{T}^{-2}\right]\left[L^3\right]}{\left[mo{l}^1\right]\left[{\theta }^1\right]}}$$
Therefore,
$$R=\left[M^1{L}^2{T}^{-2}mo{l}^{-1}{\theta }^{-1}\right]$$ …………….. (3)
Therefore, from (1), (2) and (3)
We get,
$$c=\left[M^0{L}^2{T}^{-2}{\theta }^{-1}\right]$$
$$\alpha =\left[M^0{L}^0{T}^0{\theta }^{-1}\right]$$
$$R=\left[M^1{L}^2{T}^{-2}mo{l}^{-1}{\theta }^{-1}\right]$$

Note: Dimensional analysis is the analysis of the relation between the different physical quantities by identifying their base quantities. The base quantities are mass, length, time, temperature, etc. For an equation to be correct, it should be dimensionally correct as well. Therefore, dimensional analysis is done to check the mathematical formula for various quantities. However, dimensional analysis is not the only criteria for checking the equations.