One mole of an ideal gas expands at a constan | 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 $W$ by an ideal gas is given by the formula: $W = \mu RT \ln\left(\frac{V_2}{V_1}\right)$.Given values are

Vedclass · Accepted Answer

$1728$

Vedclass · Answer

(B) For an isothermal process,the work done $W$ by an ideal gas is given by the formula: $W = \mu RT \ln\left(\frac{V_2}{V_1}ight)$.Given values are: $\mu = 1 \, mol$,$R = 8.31 \, J/mol \cdot K$,$T = 300 \, K$,$V_1 = 10 \, L$,and $V_2 = 20 \, L$.Substituting these values into the formula:$W = 1 	imes 8.31 	imes 300 	imes \ln\left(\frac{20}{10}ight)$$W = 2493 	imes \ln(2)$Using $\ln(2) \approx 0.6931$:$W = 2493 	imes 0.6931 \approx 1728 \, J$.

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

Similar Questions

$A$ perfect gas of volume $5 \ L$ is compressed isothermally to a volume of $1 \ L$. The $r.m.s.$ speed of the molecules will

For the $P-V$ diagram given for an ideal gas,which of the following correctly represents the $T-P$ diagram?

What is the value of the specific heat of a gas during an isothermal process?

$A$ gas undergoes a change at constant temperature. Which of the following quantities remain fixed?

Heat is given to an ideal gas in an isothermal process. Then

Vedclass Test Series

Exam Paper Generator

Online Exam Module