Two AIPMT 2008 Questions from Thermal Physics

The following questions appeared in the All India Pre-Medical / Pre-Dental Entrance 2008 Examination:

(1) At 10º C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110º C this ratio is

(1) 383x/283

(2) 10x/110

(3) 283x/383

(4) x

We have PV = rT where r is the gas constant for the given mass.

Since the volume V = M/ρ where M is the mass and ρ is the density, we have

P/ρ = rT/M

Since r and M are constants for a given mass of gas, we have

(ρ₁/P₁) /(ρ₂/P₂) = T₂/T₁ where the suffix 1 is for the quantities at temperature 10º C and suffix 2 is for the quantities at temperature 110º.

It is given that (ρ₁/P₁) = x.

Therefore x/(ρ₂/P₂) = 383/283 since the temperatures are in degree kelvin.

From this (ρ₂/P₂) = 283x/383

(2) On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39º W and 239º W respectively. What will be the temperature on the new scale, corresponding to a temperature of 239º C on the Celsius scale?

(1) 117º W

(2) 200º W

(3) 139º W

(4) 78º W

We have 0º C = 39º W and 100º C = 239º W

A temperature difference of 100º C is therefore equivalent to a temperature difference of 200º W. A temperature difference of 1º C is thus equivalent to a temperature difference of 2º W.

Therefore, a temperature 39º C is equal to [39 + (39×2)]º W = 117º W