Question

Method 1 of Q Temperature of two moles of a monoatomic gas is increased by 300 K in the process p prop V. (a) Find molar heat capacity of the gas in the given process. (b) Find heat given to the gas in that.

Answer 1

Answer:

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Explanation:

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Answer 2

Given :

No of moles = 2

Temperature increased = 300K

pV⁻¹= c

To find :

a) Molar heat capacity of the gas

b) Heat given to the gas

Solution :

• a)We have the equation pVˣ = c

• By comparing the Given equation with above equation we get “x= -1”

• We have the formula

       C= Cv + R/(1-x)

       C=(3/2R) + R/(1+1)   [for monoatomic gas Cv = 3R/2]

       C= 2R

• The molar heat capacity of the gas is 2R

• b) Q=ncΔt

      =2×2R×300

      =1200R

• The heat given to the gas is 1200R