What is the work done when 2 text{ mole} of a | vedclass

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

Question

(D) Given: $n = 2 	ext{ mol}$,$V_{1} = 5 	ext{ m}^{3}$,$V_{2} = 10 	ext{ m}^{3}$,$T = 300 	ext{ K}$,$R = 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.

Vedclass · Accepted Answer

$-3.458$

Vedclass · Answer

(D) Given: $n = 2 	ext{ mol}$,$V_{1} = 5 	ext{ m}^{3}$,$V_{2} = 10 	ext{ m}^{3}$,$T = 300 	ext{ K}$,$R = 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.For an isothermal reversible expansion,the work done is given by the formula: $W = -2.303 nRT \log_{10} \frac{V_{2}}{V_{1}}$.Substituting the values: $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log_{10} \frac{10}{5}$.$W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes \log_{10} 2$.Since $\log_{10} 2 \approx 0.3010$,we get: $W = -2.303 	imes 2 	imes 8.314 	imes 300 	imes 0.3010$.$W \approx -3457.97 	ext{ J} = -3.458 	ext{ kJ}$.

What is the work done when $2 \text{ mole}$ of an ideal gas are expanded isothermally and reversibly from $5 \text{ m}^{3}$ to $10 \text{ m}^{3}$ at $300 \text{ K}$ (in $\text{ kJ}$)? $(R=8.314 \text{ J K}^{-1} \text{ mol}^{-1})$

Similar Questions

Observe the following properties: Volume,enthalpy,density,temperature,heat capacity,pressure,internal energy. The number of extensive properties in the above list is

If $5 \ mol$ of an ideal gas expands from $10 \ L$ to a volume of $100 \ L$ at $300 \ K$ under isothermal and reversible conditions,then the work done,$w$,is $-x \ J$. The value of $x$ is $ . . . . . . $. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

Explain mechanical work.

Define open,closed,and isolated systems with examples.

Explain the enthalpy of dilution.

Vedclass Test Series

Exam Paper Generator

Online Exam Module