The temperature of 1 mole of an ideal monoato | vedclass

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

Question

(A) For an isobaric process,the heat added is given by $E_1 = n C_p \Delta T$. The change in internal energy is given by $E_2 = n C_v \Delta T$. There

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) For an isobaric process,the heat added is given by $E_1 = n C_p \Delta T$. The change in internal energy is given by $E_2 = n C_v \Delta T$. Therefore,the ratio is $\frac{E_1}{E_2} = \frac{n C_p \Delta T}{n C_v \Delta T} = \frac{C_p}{C_v} = \gamma$. For a monoatomic gas,the degrees of freedom $f = 3$. The adiabatic index is $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$. Given $\frac{E_1}{E_2} = \frac{x}{9}$,we have $\frac{5}{3} = \frac{x}{9}$. Solving for $x$,we get $x = \frac{5 	imes 9}{3} = 15$.

The temperature of $1$ mole of an ideal monoatomic gas is increased by $50^{\circ} C$ at constant pressure. The total heat added and change in internal energy are $E_1$ and $E_2$,respectively. If $\frac{E_1}{E_2}=\frac{x}{9}$,then the value of $x$ is . . . . . . .

Similar Questions

If $R$ is the universal gas constant,the amount of heat needed to raise the temperature of $2$ moles of an ideal monoatomic gas from $273 \ K$ to $373 \ K$ when no work is done is: (in $R$)

An amount of $700 \,J$ of heat is transferred to a diatomic gas, allowing it to expand with the pressure held constant. The work done by the gas is (in $\,J$)

In a given process for an ideal gas,$dW = 0$ and $dQ < 0$. Then for the gas:

$A$ monoatomic gas $(\gamma = 5/3)$ expands at constant pressure when heat is supplied. What percentage of the heat supplied is converted into internal energy and work, respectively?

$A$ diatomic gas with rigid molecules does $10 \ J$ of work when expanded at constant pressure. What would be the heat energy absorbed by the gas,in this process ..... $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module