Question

please answer step by step clearly:

You have 12.0 mol of a diatomic ideal gas at 300 K, and 4.00 kJ of heat to add to it. You can choose to heat it isochorically or isobarically. Which of these two methods will result in a higher final temperature, and what will that final temperature be in Kelvin?

Answer 1

So, in Isochoric process heat exchange is

AQ = mC,T -T)

and for the same heat exchange in isobaric process

AQ = mcp(T – T2)

so for diatomic gas

Cu = -R

and

R Cp = Cy + n

so,

5 R Cp= R + 2 12 = 31 -R 12

so, as

Cp > Cv that's why for the same heat exchange Temprature raised in Isobaric process is less than the isochoric process

Qܠ2 T +T 5mR

and

T2 = 12AQ 31mR +T

So that's why T2 < T1

So plug in value here T = 300K

Del Q = 4 kJ

mass is not given so take m= 1kg

R = 8.135 J/K.mol

So we get, final Temprature

T1 = 496.68 K

And

T2 = 490.33 K

So T1 > T2

So isochoric process results in higher temperature.