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Slope of isotherm for a gas (having gamm...

Slope of isotherm for a gas (having `gamma = (5)/(3)`) is `3 xx 10^(5) N//m^(2)` . If the same gas is undergoing adiabatic change then adiabatic elasticity at that instant is

A

`3 xx 10^(5) N//m^2`

B

`5 xx 10^(5) N//m^2`

C

`6 xx 10^(5) N//m^2`

D

`10 xx 10^(5) N//m^2`

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The correct Answer is:
To solve the problem step by step, we need to find the adiabatic modulus of elasticity for the gas given the slope of the isotherm and the value of gamma (γ). ### Step 1: Understand the given information We are given: - The slope of the isotherm (dP/dV) = 3 × 10^5 N/m² - The adiabatic constant (γ) = 5/3 ### Step 2: Relate the slope of the isotherm to the isothermal modulus of elasticity The slope of the isotherm is related to the isothermal modulus of elasticity (E_iso) by the equation: \[ E_{\text{iso}} = -\frac{dP}{dV} \] Since we are interested in the modulus, we take the absolute value: \[ E_{\text{iso}} = \left| -\frac{dP}{dV} \right| = \frac{dP}{dV} = 3 \times 10^5 \, \text{N/m}^2 \] ### Step 3: Use the relationship for adiabatic modulus of elasticity The adiabatic modulus of elasticity (E_adiabatic) is given by the formula: \[ E_{\text{adiabatic}} = \gamma \frac{P}{V} \] We need to find the value of P/V. From the isothermal process, we know that: \[ \frac{dP}{dV} = -\frac{P}{V} \] Thus, \[ \frac{P}{V} = -\frac{dP}{dV} \] ### Step 4: Substitute the slope into the equation Since we have the slope (dP/dV) from the isotherm: \[ \frac{P}{V} = -\left(3 \times 10^5\right) \] Now substituting this into the equation for adiabatic modulus of elasticity: \[ E_{\text{adiabatic}} = \gamma \left(-\frac{dP}{dV}\right) \] Substituting the values: \[ E_{\text{adiabatic}} = \frac{5}{3} \times \left(3 \times 10^5\right) \] ### Step 5: Calculate the adiabatic modulus of elasticity Now, simplifying the expression: \[ E_{\text{adiabatic}} = \frac{5}{3} \times 3 \times 10^5 = 5 \times 10^5 \, \text{N/m}^2 \] ### Final Answer The adiabatic modulus of elasticity at that instant is: \[ E_{\text{adiabatic}} = 5 \times 10^5 \, \text{N/m}^2 \] ---
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