For an adiabatic compression of 1 text{ kmol} | vedclass

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

Question

(B) According to the first law of thermodynamics,$\Delta Q = \Delta E_{int} + \Delta W$.For an adiabatic process,$\Delta Q = 0$.Therefore,$0 = \Delta

Vedclass · Accepted Answer

Diatomic

Vedclass · Answer

(B) According to the first law of thermodynamics,$\Delta Q = \Delta E_{int} + \Delta W$.For an adiabatic process,$\Delta Q = 0$.Therefore,$0 = \Delta E_{int} + \Delta W$,which implies $\Delta E_{int} = -\Delta W$.The change in internal energy is given by $\Delta E_{int} = n C_V \Delta T$,where $n$ is the number of moles.Given $n = 1 	ext{ kmol} = 1000 	ext{ mol}$,$\Delta W = -146 	ext{ kJ} = -146 	imes 10^3 	ext{ J}$ (work done on the gas is negative in the system's perspective for expansion,but here compression is done,so $\Delta W = -146 	ext{ kJ}$),and $\Delta T = 7 	ext{ K}$.$C_V = \frac{-\Delta W}{n \Delta T} = \frac{-(-146 	imes 10^3)}{1000 	imes 7} = \frac{146}{7} \approx 20.86 	ext{ J mol}^{-1} 	ext{ K}^{-1}$.For a diatomic gas,$C_V = \frac{5}{2} R = 2.5 	imes 8.314 \approx 20.78 	ext{ J mol}^{-1} 	ext{ K}^{-1}$.Since the calculated $C_V$ matches the value for a diatomic gas,the gas is diatomic.

For an adiabatic compression of $1 \text{ kmol}$ of gas,$146 \text{ kJ}$ of work is done. During this process,the temperature of the gas increases by $7 \text{ °C}$. The gas is ........

Similar Questions

In which of the following processes,heat is neither absorbed nor released by a system?

$A$ van der Waals gas obeys the equation of state $\left(p+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$. Its internal energy is given by $U=C T-\frac{n^2 a}{V}$. The equation of a quasistatic adiabat for this gas is given by

At $27^\circ C$,a gas is suddenly compressed such that its pressure becomes $1/8$th of the original pressure. The temperature of the gas will be $(\gamma = 5/3)$.

$A$ diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_1$ (in Kelvin) and the final temperature is $a T_1$,the value of $a$ is

An ideal gas at pressure $p$ is adiabatically compressed so that its density becomes twice that of the initial. If $\gamma = \frac{c_p}{c_v} = \frac{7}{5}$,then the final pressure of the gas is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module