An ideal gas on isothermal reversible compres | vedclass

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

Question

(C) Given: $V_1 = 10 \ L$,$V_2 = 5 \ L$,$W = 1730 \ J$,$T = 300 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.For isothermal reversible compression,the wor

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given: $V_1 = 10 \ L$,$V_2 = 5 \ L$,$W = 1730 \ J$,$T = 300 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.For isothermal reversible compression,the work done on the system is given by the formula:$W = -2.303 \ nRT \log_{10} \left(\frac{V_2}{V_1}ight)$Since work is done on the gas,$W$ is positive in this context $(W = 1730 \ J)$:$1730 = -2.303 	imes n 	imes 8.314 	imes 300 	imes \log_{10} \left(\frac{5}{10}ight)$$1730 = -2.303 	imes n 	imes 8.314 	imes 300 	imes \log_{10} (0.5)$$1730 = -2.303 	imes n 	imes 8.314 	imes 300 	imes (-0.3010)$$1730 = n 	imes 5744.14 	imes 0.3010$$1730 = n 	imes 1729$$n = \frac{1730}{1729} \approx 1 \ mol$.

An ideal gas on isothermal reversible compression from $10 \ L$ to $5 \ L$ performs $1730 \ J$ of work at $300 \ K$. Calculate the number of moles of gas involved in the compression? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

Similar Questions

$A$ gas occupies $0.418 \, L$ at $27 \, ^oC$ and $740 \, mm \, Hg$ pressure. If the mass of the gas is $3.0 \, g$,what is its molar mass?

If $R = 8.314$,express it with its units.

The mathematical form of Charles's law is given by $V_{t} = V_{0} \left( \frac{273.15 + t}{273.15} \right)$. Using this equation,explain the relationship between volume $(V)$ and temperature $(T)$ in Kelvin.

Which of the following is equal to $[{\partial P/\partial V}]_T$ for an ideal gas?

Molecular attraction and size of the molecules in a gas are negligible at :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module