A monoatomic gas is taken through a process T | vedclass

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

Question

(C) Given the process equation: $TP^{-1/3} = 	ext{constant}$.Using the ideal gas law $PV = nRT$, we have $T \propto PV$, so $PV \cdot P^{-1/3} = 	ex

Vedclass · Accepted Answer

Its temperature will decrease

Vedclass · Answer

(C) Given the process equation: $TP^{-1/3} = 	ext{constant}$.Using the ideal gas law $PV = nRT$, we have $T \propto PV$, so $PV \cdot P^{-1/3} = 	ext{constant}$, which simplifies to $P^{2/3}V = 	ext{constant}$.Raising both sides to the power of $3/2$, we get $PV^{3/2} = 	ext{constant}$.This is a polytropic process $PV^x = 	ext{constant}$ where $x = 3/2$.For a monoatomic gas, the adiabatic index $\gamma = 5/3$.The molar heat capacity $C$ for a polytropic process is given by $C = \frac{R}{\gamma - 1} + \frac{R}{1 - x}$.Substituting the values: $C = \frac{R}{5/3 - 1} + \frac{R}{1 - 3/2} = \frac{R}{2/3} + \frac{R}{-1/2} = \frac{3R}{2} - 2R = -\frac{R}{2}$.Since the molar heat capacity $C$ is negative, from the relation $\Delta Q = nC\Delta T$, if heat is given to the gas $(\Delta Q > 0)$, then $\Delta T$ must be negative $(\Delta T Therefore, the temperature of the gas will decrease.

$A$ monoatomic gas is taken through a process $TP^{-1/3} = \text{constant}$. If heat is given to the gas, what happens to its temperature?

Similar Questions

$2$ moles of a monoatomic gas are expanded to double its initial volume, through a process $P/V = \text{constant}$. If its initial temperature is $300\, K$, then which of the following is not true?

$A$ gas expands with temperature according to the relation $V = k T^{2/3}$. What is the work done when the temperature changes by $30^{\circ}C$ (in $R$)?

$A$ monatomic ideal gas undergoes a process in which the ratio of $P$ to $V$ at any instant is constant and equals to $1$. What is the molar heat capacity of the gas?

$A$ gas obeys $P^2V = \text{constant}$. The initial temperature and volume are $T_0$ and $V_0$. If the gas expands to a volume of $2V_0$,the final temperature is:

$2$ moles of a diatomic gas undergoes the process: $PT^2/V = \text{constant}$. The molar heat capacity of the gas during the process is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module