One mole of an ideal gas expands from an init | vedclass

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

Question

(B) For an isothermal process,the work done by an ideal gas is given by the formula: $W = \mu RT \ln\left(\frac{V_2}{V_1}\right)$ Given: Number of mol

Vedclass · Accepted Answer

$1728$

Vedclass · Answer

(B) For an isothermal process,the work done by an ideal gas is given by the formula: $W = \mu RT \ln\left(\frac{V_2}{V_1}ight)$ Given: Number of moles $\mu = 1 \; mol$ Gas constant $R = 8.31 \; J/mol \cdot K$ Temperature $T = 300 \; K$ Initial volume $V_1 = 10 \; L$ Final volume $V_2 = 20 \; L$ Substituting the values: $W = 1 	imes 8.31 	imes 300 	imes \ln\left(\frac{20}{10}ight)$ $W = 2493 	imes \ln(2)$ Using $\ln(2) \approx 0.693$: $W \approx 2493 	imes 0.693 \approx 1727.65 \; J$ Rounding to the nearest integer,we get $W \approx 1728 \; J$.

One mole of an ideal gas expands from an initial volume of $10 \; L$ to a final volume of $20 \; L$ at a constant temperature of $300 \; K$. Calculate the work done by the gas in $J$. $(R = 8.31 \; J/mol \cdot K)$ (in $; J$)

Similar Questions

One mole of ${O_2}$ gas having a volume equal to $22.4 \text{ litres}$ at $0^{\circ}C$ and $1 \text{ atmospheric pressure}$ is compressed isothermally so that its volume reduces to $11.2 \text{ litres}$. The work done in this process is ...... $J$.

An isothermal process has a non-linear graph between:

An ideal gas is subjected to an isothermal expansion such that its volume changes from $V_i$ to $V_f$ and pressure from $P_i$ to $P_f$. The work done on the gas is:

For an isothermal expansion of a perfect gas,the value of $\frac{\Delta P}{P}$ is equal to:

Write the first law of thermodynamics for an isothermal process in an ideal gas.

Vedclass Test Series

Exam Paper Generator

Online Exam Module