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Marrrta [24]
3 years ago
5

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 p

ressure increases by 50 mb. What is the change in the temperature of the air and what is the amount of heat absorbed?
Physics
1 answer:
Rufina [12.5K]3 years ago
7 0

Answer:

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

<u>ΔT = 13.65° C</u>

<u></u>

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)

<u>ΔQ = 13.7 J</u>

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