An ideal gas has molar heat capacity `C_v` at constant volume. The gas undergoes a process where in the temperature changes as `T= T_0 (1+ alpha V^2)`, where T and V are temperature and volume respectively, `T_0` and a are positive constants. The molar heat capacity C of the gas is given as `C = C_v + R f(v)`, where `f(v)` is a function of volume. The expression for `f(v)` is

Options

A. `(alpha V^2)/(1 + alpha V^2)`
B. `(1 + alpha V^2)/(2 alpha V^2)`
C. `( alpha V^2 ( 2+ alpha V^2)`
D. `(1)/( 2 alpha V^2 ( 1+ alpha V^2))`

Answer

B

Explanation

Given,
For ideal gas,
Molar heat capacity at constant volume = `C_v`
Temperature, `T = T_0 ( 1+ alpha V_2) " "…(1)`
Where,
V - Volume of gas,
`T _(0), alpha - "positive constants `
Molar heat capacity of the gas , `C = C_(v + Rg (V )" "(2)`
Where,
`F (V) -` Function of volume. But the molar
Heat capacity is given by the equation,
`C = (d Q)/(d T)" "...(3)`
According to first law of thermodynamics,
`d Q = dU + dW`
`= C_(v) dT + pd V`
`rArr d Q = C_(v) d T + (RT)/( V) d V ( because PV = RT)" "...(4)`
From equation (1),
`T = T_(0) ( 1 + alpha V^2)`
`T = T_(0) + alpha T_(0) V^2`
Differentiating on both sides,
`rArr dT = 0 + alpha T_0 2 V d V`
`rArr dT = 2 alpha T_(0) V dV`
`rArr dV = (dT)/(2 alpha T_(0) V )" "...(5)`
Substituting equation (5) in equation (4)
`rArr d Q = C_(v) d T + (RT)/(V ) xx (d T)/(2 alpha T_(0) V )`
`rArr d Q = (C_(v) = (RT)/(2 alpha T_(0) V^2))dT`
`rArr (dQ)/(dT) = (C_(v) + (RT)/(2 alpha T_(0) V^2))`
`rArr C= C_(v) + (RT)/(2 alpha T_(0) V^2) " "..(6) (because "From equation"(1))`
Equating equation (6) & (2)
`rArr C_(v) + R r (V) = C_(v) + (RT)/(2 alpha T_(0) V^2)`
`Rf (V) = (RT)/(2 alpha T_(0) V^2)`
`f(V) = (T)/(2 alpha T_(0) V^2)`
`f(V) = (T_(0) (1 + alpha V^2)/(2 alpha T_(0) V^2)) (because " From equation"(1))`
`f(V) = (1+ alpha V^2)/( 2 alpha V^2)`
` therefore f(V) = (1 + alpha V^2)/(2 alpha V^2)`