An ideal gas expands isothermally and reversi | vedclass

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

Question

(B) For an isothermal reversible expansion of an ideal gas,the work done by the gas is given by: $W = nRT \ln\left(\frac{V_2}{V_1}\right)$ Given: $W =

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) For an isothermal reversible expansion of an ideal gas,the work done by the gas is given by: $W = nRT \ln\left(\frac{V_2}{V_1}ight)$ Given: $W = 5.187 \ kJ = 5187 \ J$ $V_1 = 10 \ m^3$ $V_2 = 20 \ m^3$ $T = 300 \ K$ $R = 8.314 \ J \ mol^{-1} \ K^{-1}$ Substituting the values: $5187 = n 	imes 8.314 	imes 300 	imes \ln\left(\frac{20}{10}ight)$ $5187 = n 	imes 8.314 	imes 300 	imes 0.6931$ $5187 = n 	imes 1728.5$ $n = \frac{5187}{1728.5} \approx 3$ Therefore,the number of moles of gas used is $3$.

An ideal gas expands isothermally and reversibly from $10 \ m^3$ to $20 \ m^3$ at $300 \ K$,performing $5.187 \ kJ$ of work on the surroundings. Calculate the number of moles of gas used.

Similar Questions

At what temperature will the volume of a gas be halved if the initial temperature is $17^{\circ}C$ and the pressure remains constant (in $^{\circ}C$)?

If $2$ moles of an ideal gas at $546 \ K$ has a volume of $44.8 \ L$,then what will be its pressure (in $atm$)? $(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1})$

What does the following graph represent?

$A$ certain mass of a gas occupies a volume of $2 \ dm^{3}$ at $STP$. At what temperature will the volume of the gas become double,keeping the pressure constant (in $^{\circ} C$)?

An open vessel containing air was heated from $27^{\circ} C$ to $727^{\circ} C$. Some air was expelled. What is the fraction of air remaining in the vessel? (Assume air as an ideal gas)

Vedclass Test Series

Exam Paper Generator

Online Exam Module