At T = 0C and p = 1000mb, 1 g of dry air receives an amount of heat during an isochoric process. It is then observed that its pressure increases by 50 mb. What is the change in the temperature of the air and what is the amount of heat absorbed?

ΔT = 13.65° C

ΔQ = 13.7 J

Explanation:

First we will find the final temperature of air by using equation of state:

P₁V₁/T₁ = P₂V₂/T₂

For Isochoric Process, V₁ = V₂

Therefore,

P₁/T₁ = P₂/T₂

T₂ = P₂T₁/P₁

where,

T₂ = Final Temperature = ?

P₂ = Final Pressure = 1050 mb

P₁ = Initial Temperature = 1000 mb

T₁ = Initial Temperature = 0°C = 273 k

Therefore,

T₂ = (1050 mb)(273 K)/(1000 mb)

T₂ = 286.65 K

Change in Temperature = ΔT = T₂ - T₁

ΔT = 286.65 K - 273 K

ΔT = 13.65° C

The first law of thermodynamics can be written as:

ΔQ = ΔU + W

where,

ΔQ = heat absorbed

ΔU = change in internal energy = mCΔT

W = Work Done = 0 (in case of isochoric process)

Therefore.

ΔQ = mCΔT

where,

m = mass of air = 1 g = 1 x 10⁻³ kg

C = specific heat of dry air = 1003.5 J/kg.°C

Therefore,

ΔQ = (1 x 10⁻³ kg)(1003.5 J/kg.°C)(13.65°C)

ΔQ = 13.7 J

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