Question

A sample of gas expand isothermally from 10L to 20L against a pressure 2 atm. Calculate the work done in tye process.

Answer 1

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Answer 2

Answer:

The work done by a sample of gas from$10L$ to $20L$ against a pressure of $2atm$ is $13.86 joule$

Explanation:

The equation used to calculate the work done in the isothermal process

                            $W = -nRT ln\frac{V_{2} }{V_1}$  = $-2.303nRT log\frac{V_{2} }{V_{1} }$

we know that $PV = nRT$

So, the equation will be $W =$ $-2.303 P_1V_1 log\frac{V_2}{V_1}$

Given that,

$P_1 = 10L\\ V_1 = 10L\\ V_2 = 20L$

$W = -2.303$×$10L$×$2 atm$×$log\frac{20}{10}$

$W = -2.303$×$10L$×$2 atm$×$0.301$

$W = 13.86 joule$

The work done by a sample of gas from$10L$ to $20L$ against a pressure of $2atm$ is $13.86 joule$