Consider the figure provided. 1 mol of an id | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a reversible isothermal expansion,the work done by the system is given by: $\omega = -nRT \ln \left(\frac{V_2}{V_1}\right)$.Given:$n = 1 \ mol

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(B) For a reversible isothermal expansion,the work done by the system is given by: $\omega = -nRT \ln \left(\frac{V_2}{V_1}ight)$.Given:$n = 1 \ mol$$T = 18 + 273.15 = 291.15 \ K$$V_1 = 10 \ L$$V_2 = 10 + 90 = 100 \ L$$R = 0.08206 \ L \ atm \ mol^{-1} \ K^{-1}$Substituting the values:$\omega = -1 	imes 0.08206 	imes 291.15 	imes \ln \left(\frac{100}{10}ight)$$\omega = -23.887 	imes \ln(10)$$\omega = -23.887 	imes 2.303$$\omega \approx -55.01 \ L \ atm$The magnitude of work done by the system is $55 \ L \ atm$.

Similar Questions

The density of neon will be highest at

The root mean square velocity of an ideal gas in a closed container of fixed volume is increased from $5 \times 10^4 \ cm \ s^{-1}$ to $10 \times 10^4 \ cm \ s^{-1}$. Which of the following statements correctly explains how the change is accomplished?

For an ideal gas obeying Boyle's Law,which of the following is correct? ($K$ is a constant)

At what pressure will a quantity of gas,which occupies $100 \ mL$ at a pressure of $720 \ mm$,occupy a volume of $84 \ mL$?

The correct combined gas law equation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module