One mole of an ideal gas is taken through an | vedclass

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

Question

(A) For a polyatomic gas,the degrees of freedom $f$ is calculated as: $f = f_{	ext{trans}} + f_{	ext{rot}} + f_{	ext{vib}}$.Given $f_{	ext{trans}}

Vedclass · Accepted Answer

Work done on the gas is close to $582\,J$

Vedclass · Answer

(A) For a polyatomic gas,the degrees of freedom $f$ is calculated as: $f = f_{	ext{trans}} + f_{	ext{rot}} + f_{	ext{vib}}$.Given $f_{	ext{trans}} = 3$,$f_{	ext{rot}} = 3$,and $f_{	ext{vib}} = 2 	imes 4 = 8$ (since each vibrational mode contributes $2$ degrees of freedom).Thus,$f = 3 + 3 + 8 = 14$.The adiabatic index $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{14} = 1 + \frac{1}{7} = \frac{8}{7}$.The work done in an adiabatic process is given by $W = \frac{nR\Delta T}{1-\gamma}$.Here,$n = 1$,$R = 8.314\,J/mol\cdot K$,$\Delta T = 37 - 27 = 10\,K$,and $\gamma = 8/7$.$W = \frac{1 	imes 8.314 	imes 10}{1 - 8/7} = \frac{83.14}{-1/7} = -83.14 	imes 7 = -581.98\,J \approx -582\,J$.Since the work done $W$ is negative,work is done on the gas.

One mole of an ideal gas is taken through an adiabatic process where the temperature rises from $27^{\circ}C$ to $37^{\circ}C$. If the ideal gas is composed of polyatomic molecules that have $4$ vibrational modes,which of the following is true?

Similar Questions

$A$ polyatomic gas $\left( \gamma = \frac{4}{3} \right)$ is compressed to $\frac{1}{8}$ of its volume adiabatically. If its initial pressure is $P_0$,its new pressure will be (in $P_0$)

What is the name of the ideal-gas process in which no heat is transferred?

$A$ monoatomic gas is suddenly compressed to $(1/8)^{\text{th}}$ of its initial volume adiabatically. The ratio of the final pressure to initial pressure of the gas is $(\gamma = 5/3)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module