Question

the volume expansion coefficient of an ideal gas Gama at a constant pressure is observed as gamma directly proportional T^n the value of n is
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Answer 1

Answer:

At a constant pressure is observed as gamma directly proportional T^(n) the value of n is -1

Explanation:

As we know that when gas is held at constant pressure then we have

$V = \frac{nR}{P} T$

here we know that

$\frac{nR}{P} = constant$

now we have

$\frac{dV}{dT} = \frac{nR}{P}$

now we know that volume expansion coefficient is given as

$\gamma = \frac{1}{V}(\frac{dV}{dT})$

so we have

$\gamma = \frac{1}{T}$

so value of n = -1

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Answer 2

Gamma is directly proportional T^n where n is equal to -1.

Explanation:

The coefficient of volume expansion of a gas at constant pressure is defined as the fraction of its volume at 0°C by which the volume of a fixed mass of gas expands per degree Celsius rise in temperature.It is given by :

$\gamma=\dfrac{1}{V}\dfrac{dV}{dT}$..........(1)

Since,

PV = nRT

$V=\dfrac{nRT}{P}$..........(2)

$\Delta V=\dfrac{nR\Delta T}{P}$..............(3)

Dividing equation 2 and 3 we get :

$\dfrac{\Delta V}{V}=\dfrac{\Delta T}{T}$

$\Delta V=\dfrac{\Delta TV}{T}$

So,

$\gamma=\dfrac{1}{T}$

i.e. n = -1

So, gamma is directly proportional T^n where n is equal to -1. Hence, this is the required solution.

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