6 moles of an ideal gas expand isothermally a | vedclass

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

Question

(D) The work done in an isothermal reversible expansion is given by the formula: $W = -nRT \ln(\frac{V_2}{V_1})$Using the base-$10$ logarithm: $W = -2

Vedclass · Accepted Answer

$34.465$

Vedclass · Answer

(D) The work done in an isothermal reversible expansion is given by the formula: $W = -nRT \ln(\frac{V_2}{V_1})$Using the base-$10$ logarithm: $W = -2.303 	imes nRT \log(\frac{V_2}{V_1})$Given values: $n = 6 \, mol$,$R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$,$T = 27 + 273 = 300 \, K$,$V_1 = 1 \, L$,$V_2 = 10 \, L$.Substituting the values: $W = -2.303 	imes 6 	imes 8.314 	imes 300 	imes \log(\frac{10}{1})$$W = -2.303 	imes 6 	imes 8.314 	imes 300 	imes 1$$W = -34464.8 \, J = -34.465 \, kJ$The magnitude of the maximum work done is $34.465 \, kJ$.

$6$ moles of an ideal gas expand isothermally and reversibly from a volume of $1$ litre to a volume of $10$ litres at $27\,^oC$. What is the maximum work done (in $, kJ$)?

Similar Questions

Explain extensive property and intensive property.

Out of $E, H, G, q$ and $w$,which are state functions?

An adiabatic process is experienced by a/an ...... system.

Which of the following is not an intensive property?

$5 \ mol$ of an ideal monoatomic gas is compressed isothermally from $(40 \ L, 5 \ bar)$ to $4 \ L$. The magnitude of maximum work done on the gas in this compression is

Vedclass Test Series

Exam Paper Generator

Online Exam Module