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One mole of a monatomic ideal gas underg...

One mole of a monatomic ideal gas undergoes a quasistatic process, which is depicted by a straight line joining points `(V_(0), T_(0))` and `(2V_(0), 3T_(0))` in a V-T diagram. What is the value of the heat capacity of the gas at the point `(V_(0), T_(0))` ?

A

R

B

`3/2 R`

C

`2R`

D

`0`

Text Solution

Verified by Experts

The correct Answer is:
C

`dQ = dU + pdV`
or, `dQ = nC_(v) T + pdV`
or, `dQ = C_(v) dT + pdV` [given n = 1 mol] ….(1)
For monatomic gas,
`gamma = 5/3 or, (C_p)/(C_v) = 5/3 or, C_(p) = 5/3 C_(v)`
Now, `C_(p) - C_(v) = R`
or, `5/3 C_(v) - C_(v) = R or, C_(v) = 3/2 R`
`:.` From (1) we get,
`dQ = 3/2 R xx (3T_(0) -T_(0)) + (RT_0)/(V_0) xx (2V_(0) - V_(0))`
or, `dQ = 4 RT_(0)`
or, `CdT = 4 RT_(0)` [C is heat capacity]
or, `C xx 2T_(0) = 4RT_(0) or, C = 2R`.
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