The figure shows the variation in heat suppli | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For an isobaric process,the heat supplied is given by $Q = n C_p \Delta T$,where $n$ is the number of moles and $C_p$ is the molar heat capacity a

Vedclass · Accepted Answer

$P, D$ and $M$

Vedclass · Answer

(A) For an isobaric process,the heat supplied is given by $Q = n C_p \Delta T$,where $n$ is the number of moles and $C_p$ is the molar heat capacity at constant pressure.From the graph,the slope of the line is $\frac{Q}{\Delta T} = n C_p$.Since the number of moles $n$ is the same for all gases,the slope is directly proportional to $C_p$.The molar heat capacity at constant pressure is $C_p = C_v + R = \left( \frac{f}{2} + 1 ight) R$,where $f$ is the degree of freedom.For monoatomic gas $(M)$,$f = 3$,so $C_p = (1.5 + 1) R = 2.5 R$.For diatomic gas $(D)$,$f = 5$,so $C_p = (2.5 + 1) R = 3.5 R$.For polyatomic gas $(P)$,$f = 6$,so $C_p = (3 + 1) R = 4 R$.Thus,$C_p(P) > C_p(D) > C_p(M)$.Since the slope is proportional to $C_p$,the slopes follow the order: $	ext{slope}(a) > 	ext{slope}(b) > 	ext{slope}(c)$.Therefore,line $a$ corresponds to $P$,line $b$ corresponds to $D$,and line $c$ corresponds to $M$.

Similar Questions

The molar specific heat of an ideal gas at constant pressure and constant volume is $C_p$ and $C_v$ respectively. If $R$ is the universal gas constant and the ratio of $C_p$ to $C_v$ is $\gamma$,then $C_v=$

The ratio of specific heats $\left(\frac{C_{P}}{C_{V}}\right)$ in terms of degree of freedom $(f)$ is given by

If a gas has $n$ degrees of freedom, then the ratio of $\frac{C_p}{C_V}$ is

The specific heat at constant pressure and at constant volume for an ideal gas are $C_p$ and $C_v$,and its adiabatic and isothermal elasticities are $E_{\phi}$ and $E_{\theta}$ respectively. The ratio of $E_{\phi}$ to $E_{\theta}$ is

One mole of an ideal monoatomic gas is heated at a constant pressure from $0^{\circ} C$ to $100^{\circ} C$. The change in the internal energy of the gas is (Given, $R = 8.32 \text{ J mol}^{-1} \text{ K}^{-1}$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module