Question

Find the equation for the work of a reversible, isothermal compression of 1 mol of gas in a pistonlcylinder assembly if the molar volume of the gas is given by where b and r are positive constants.

Answer 1

V= ((RT)/P) + b where b and R are positive constants 

 P = RT/(V-b) 
 plug it into the equation W = - (integral from V1 to V2) P dv

 
W = -∫PdV, from V1->V2 => -∫[RT/(V-b)]dV = -RT*ln|V-b|, from V1->V2 

W = RT[ ln|V1-b| - ln|V2-b|] = RT[ ln|(V1-b)/(V2-b)]

HOPE THIS HELPS

Answer 2

The equation for the work of a reversible, isothermal compression of 1 mol of gas in a piston-cylinder assembly, given that the molar volume of the gas is $W = -nRT \:ln[\frac{(bV_f + \frac{V_f^2}{r} )}{(bV_i + \frac{V_i^2}{r} )} ]$

The following equation can be used to determine the amount of work produced by a reversible, isothermal compression of a gas:

$W = -nRT \:ln(V_{f} /V_{i} )$

Where:

W is the work done

n is the value of number of moles of gas present

R is the gas constant

T is the temperature of the gas

$V_i$ is the initial gas volume

$V_{f}$ is the final gas volume

In this instance, we are informed that the gas's molar volume is determined by:

V = b + $\frac{r}{p}$

Where:

V is the molar volume of the gas

P is the pressure of the gas

We must first describe the gas pressure as a function of the volume of the gas in order to solve for the work done.

To do this, we may rewrite the molar volume equation as follows:

P = $\frac{r}{(V - b)}$

Now, we can add this expression for P to the work equation:

$W = -nRT \:ln[\frac{(b + \frac{V_f}{r} )}{(b + \frac{V_i}{r} )} ]$

By simplifying this expression, we get:

$W = -nRT \:ln[\frac{(bV_f + \frac{V_f^2}{r} )}{(bV_i + \frac{V_i^2}{r} )} ]$

Therefore, the equation for the work of a reversible, isothermal compression of 1 mol of gas in a piston-cylinder assembly, given that the molar volume of the gas is V = b + (r / P):

$W = -nRT \:ln[\frac{(bV_f + \frac{V_f^2}{r} )}{(bV_i + \frac{V_i^2}{r} )} ]$

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