Question

It takes 200 J of work to compress quasi-statically 0.40 mol of an ideal gas to one-fourth its original volume. Calculate the temperature of the gas (in K), assuming it remains constant during the compression.

Answer 1

Answer : The temperature of the gas is, 43.4 K

Explanation :

The expression used for work done will be,

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = 200 J

n = number of moles of gas  = 0.40 mol

R = gas constant = 8.314 J/mole K

T = temperature of gas  = ?

$V_1$ = initial volume of gas  = X

$V_2$ = final volume of gas  = $\frac{X}{4}$

Now put all the given values in the above formula, we get :

$200J=-0.40mol\times 8.314J/moleK\times T\times \ln (\frac{(\frac{X}{4})}{X})$

$T=43.4K$

Thus, the temperature of the gas is, 43.4 K