If the given graph shows the logarithmic valu | vedclass

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

Question

(C) For an adiabatic process, the relation between pressure $(P)$ and volume $(V)$ is given by $PV^{\gamma} = 	ext{constant}$.Taking the logarithm on

Vedclass · Accepted Answer

$1.4$

Vedclass · Answer

(C) For an adiabatic process, the relation between pressure $(P)$ and volume $(V)$ is given by $PV^{\gamma} = 	ext{constant}$.Taking the logarithm on both sides, we get $\log P + \gamma \log V = 	ext{constant}$, which can be written as $\log P = -\gamma \log V + C$.This is the equation of a straight line $y = mx + c$, where the slope $m = -\gamma$.From the graph, the slope is calculated as:$m = \frac{\log P_2 - \log P_1}{\log V_2 - \log V_1} = \frac{2.20 - 2.48}{1.4 - 1.2} = \frac{-0.28}{0.2} = -1.4$.Since $m = -\gamma$, we have $-\gamma = -1.4$, which implies $\gamma = 1.4$.Therefore, the ratio of the specific heat capacities is $1.4$.

If the given graph shows the logarithmic values of pressure $(P)$ and volume $(V)$ of an ideal gas, then the ratio of the specific heat capacities of the gas is

Similar Questions

$A$ diatomic gas undergoes adiabatic change. Its pressure $P$ and temperature $T$ are related as $P \propto T^{x}$ where the value of $x$ is

During a thermodynamic process, the increase in internal energy of a system is equal to the work done on the system. Which process does the system undergo?

During an adiabatic process,the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of $\frac{C_P}{C_V}$ for the gas is

$\Delta U + \Delta W = 0$ is valid for

One mole of helium is adiabatically expanded from its initial state $({P_i}, {V_i}, {T_i})$ to its final state $({P_f}, {V_f}, {T_f})$. The decrease in the internal energy associated with this expansion is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module