Solveeit Logo
Login

Question

Question: For a gas \(\left( \frac{dE}{dV} \right)_{T}\) = 0 then...

For a gas (dEdV)T\left( \frac{dE}{dV} \right)_{T} = 0 then

A

(dHdP)T\left( \frac{dH}{dP} \right)_{T}= P(dVdP)T\left( \frac{dV}{dP} \right)_{T} + V

B

If the gas is ideal then (dHdP)T\left( \frac{dH}{dP} \right)_{T}= 0

C

(dCvdV)T\left( \frac{dC_{v}}{dV} \right)_{T} = 0

D

All of these

Answer

All of these

Explanation

Solution

H = E + PV

(dHdP)T\left( \frac{dH}{dP} \right)_{T}= (dEdP)T\left( \frac{dE}{dP} \right)_{T}+ P(dVdP)T\left( \frac{dV}{dP} \right)_{T}+ V

or (dHdP)T\left( \frac{dH}{dP} \right)_{T} =(dVdP)T\left( \frac{dV}{dP} \right)_{T}+ P (dVdP)T\left( \frac{dV}{dP} \right)_{T}+ V

= P(dVdP)T\left( \frac{dV}{dP} \right)_{T}+ V

If the gas is ideal then PV = RT

or V =RTP\frac{RT}{P}

or (dVdP)T\left( \frac{dV}{dP} \right)_{T}= −RTP2\frac{- RT}{P^{2}}

or P(dVdP)T\left( \frac{dV}{dP} \right)_{T}= −RTP\frac{- RT}{P} = −PVP\frac{- PV}{P} = – V

and hence for ideal gas (dHdP)T\left( \frac{dH}{dP} \right)_{T} = 0

Cv = (dEdT)V\left( \frac{dE}{dT} \right)_{V}

or = ddV\frac{d}{dV}

= ddT\frac{d}{dT} {(dEdV)T}\left\{ \left( \frac{dE}{dV} \right)_{T} \right\} = 0

Because E is a state function