Two moles of an ideal mono-atomic gas undergoes a thermodynamic process in which the molar heat capacity `'C'` of the gas depends on absolute temperature as `C=(RT)/(T_(0))`, where `R` is gas consant and `T_(0)` is the initial temperature of the gas. (`V_(0)` is the initial of the gas). Then answer the following questions:
The equation of process is

Options

A. `1/P=(V_(0)T_(0)^(3//2))/(4RT^(5/2))e^((T-T_(0)/(T_(0)))`
B. `1/P=(V_(0)T_(0)^(3//2))/(2RT^(5/2))e^((T-T_(0)/(T_(0)))`
C. `1/P=(V_(0)T_(0)^(3//2))/(4RT^(5/2))e^((T+T_(0)/(T_(0)))`
D. `1/P=(V_(0)T_(0)^(3//2))/(3RT^(5/2))e^((T-T_(0)/(T_(0)))`

Answer

B

Explanation

`dQ=dU+dW`
`impliesnCdT=nC_(v)dT=PdV`
`implies(2RT)/(T_(0))-3R=(2RT)/V.(dV)/(dT)`……(1)
From (1) `int_(T_(0))^(T)(1/(T_(0))-3/(2T))dT=int_(v_(0))^(v)=(dV)/V`
On solving `V=V_(0)((T_(0))/T)^(3//2)e^(((T-T_(0))/(T_(0))))`
`1/P=(V_(0)T_(0)^(3//2))/(2RT^(5//2))e^(((T-T_(0))/(T_(0))))`
For minimum volume `(dv)/(dT)=0`
So, `T=3/2T_(0)`
`V_("min")=V_(0)((T_(0))/(3/2T_(0)))^(3//2)e^((((3)/(2)T_(0)-T_(0))/(T_(0))))`
`V_(min)=((2)/(3))^(3//2)V_(0)e^(1//2)`