One mole of the ideal gas through the process ` p= p_0 [ 1 - alpha ((V)/(V_0))^3 ]` , where p and V are pressure and volume `p_0 , V_0` and `alpha` are constant . If the maximum attainable temperature of the gas is `(3/4) (p_0 V_0)/(R ) ` , then the value of `alpha ` is

Options

A. 2
B. `1/2`
C. `1/4`
D. `4`

Answer

C

Explanation

For 1 mole of an ideal gas ,
` p = (RT)/(V)`
But ` p = p_0 (1 - (alpha V^3)/(V_0^3) )` (given )
so , `(RT)/(V) = p_0- (alpha_0 )/(V_0) V^3`
` T = (p_0 V)/(R )- (alpha p_0 V^4)/(V_0^3 R) ` ….(i)
For maximum T,
` = (dT)/(dV) = 0`
` rArr (dT)/(dV) = (p_0)/(R) - (4 alpha p_0 )/(RV_0^3) V^3 rArr (p_0)/(R) = (4alpha p_0)/(RV_0^3) V^3`
` rArr V^3 - (V_0^3)/(4 alpha ) rArr V = (V_0)/( ( 4 alpha)^(1//3) )`
Substituting in Eq. (i) , we get
`T_(max) = (p_0 V_0)/(R (4 alpha)^(1//3) ) - (alpha p_0 V_0^4)/(V_0^3 R (4 alpha)^(4//3)) = (p_0V_0)/(R(4 alpha)^(1//3) ) - (p_0 V_0)/(R4^(4//3) alpha^(1//3) )`
Substituting , ` T_(max) = (3p_0 V_0)/(4R) `, we get
`3/4 = (1)/(4^(1//3) alpha^(1//3) ) - (1)/(4^(4//3) alpha^(1//3) )`
Now from options, we can check that only `alpha = 1/4` balances the equation .
so , `alpha = 1/4`