Question
Question: For a gaseous reaction \[A\left( g \right)\to \text{Product}\], which one of the following is correc...
For a gaseous reaction A(g)→Product, which one of the following is correct relation among dtdp,dtdn and dtdc
& \dfrac{dp}{dt}=\text{Rate of relation in atm se}{{\text{c}}^{-1}} \\\ & \dfrac{dc}{dt}=\text{Rate of relation in molarity se}{{\text{c}}^{-1}} \\\ & \dfrac{dn}{dt}=\text{ Rate of reaction in mol se}{{\text{c}}^{-1}} \\\ \end{aligned} \right)$$ A. $$\dfrac{dc}{dt}=\dfrac{dn}{dt}=\dfrac{-dp}{dt}$$ B. $$\dfrac{-dc}{dt}=\dfrac{1}{v}\dfrac{dn}{dt}=\dfrac{-1}{RT}\dfrac{dp}{dt}$$ C. $$\dfrac{dc}{dt}=\dfrac{v}{RT}\dfrac{dn}{dt}=\dfrac{dp}{dt}$$. D. None of theseSolution
Hint: We can solve these type of question by considering the equation of state for an ideal gas which is given as PV=nRT
We can differentiate the equation with respect to time and find the relation between the quantities which are asked also. Also we must know that
Concentration=Volume of gasNumber of moles.
Complete answer:
From the equation of the state which is PV=nRT
Where P = Pressure exerted by the gas
V = Volume of gas
n = moles of gas
R = universal gas constant
T = Temperature
PV=nRT (differentiating the equation on both sides we get)
dtdp.V=RT.dtdn
⇒V1dtdn=RT1dtdp ........... 1
Now Consider, PV=nRT or P=VnRT
Concentration =Vn=C
Substituting this value in as one equation we get, P=CRT
Differentiating with respect to dt on both side we get,
dtdp=dtdcRT
dtdc=RT1dtdp
Hence, the correct relationship between the given value are:
dt−dc=v−1dtdn=RT−1dtdp
Hence correct option in option B.
Note: In thermodynamics the equation of state is a thermodynamic equation relating state variables which describe the state of matter under a given set of physical conditions, such as pressure volume and temperature or internal energy as and when required.