Sol:

For isobaric process,

[latex]C_{p} = \frac{\Delta Q}{n\Delta T}[/latex] [latex]C_{p} = \frac{70}{2 * (35 – 30)}[/latex] [latex]C_{p} = \frac{70}{2 * 5}[/latex] [latex]C_{p} = \frac{70}{10} = 7 calories mol^{-1}K^{-1}[/latex]

For an idea; gas,

[latex]C_{p} – C_{v} = R = 8.314J-mol^{-1}K^{-1}\simeq 2 calories mol^{-1}K^{-1}[/latex] [latex]\Rightarrow C_{v} = C_{p} – R[/latex] [latex]\Rightarrow C_{v} = (7 – 2) [/latex] [latex]\Rightarrow C_{v} = 5 calories mol^{-1}K^{-1}[/latex] [latex]\Rightarrow C_{v} = \frac{\Delta Q}{n\Delta T}[/latex] [latex]\Rightarrow 5 = \frac{\Delta Q}{2 * (35 – 30)}[/latex] [latex]\Rightarrow \Delta Q = 5 * 2 * (35 – 30)[/latex] [latex]\Rightarrow \Delta Q = 5 * 2 * 5[/latex] [latex]\Rightarrow \Delta Q = 50[/latex]

Therefore, 50 calories of heat are required to raise the temperature of two moles of

gas from 30° – 35° C at a constant volume.

Explore more such questions and answers at BYJU’S.

Was this answer helpful?

 
   

0 (0)

(0)
(0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question