If 2.5 moles of an ideal gas at a certain tem | vedclass

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

Question

(C) The formula for reversible isothermal work is: $W = -2.303 nRT \log \left(\frac{V_2}{V_1}\right)$Given: $n = 2.5 \ mol$, $V_1 = 2 \ dm^3$, $V_2 =

Vedclass · Accepted Answer

$345$

Vedclass · Answer

(C) The formula for reversible isothermal work is: $W = -2.303 nRT \log \left(\frac{V_2}{V_1}ight)$Given: $n = 2.5 \ mol$, $V_1 = 2 \ dm^3$, $V_2 = 20 \ dm^3$, $W = -16.5 \ kJ = -16500 \ J$, $R = 8.314 \ J \ K^{-1} \ mol^{-1}$Substituting the values: $-16500 = -2.303 	imes 2.5 	imes 8.314 	imes T 	imes \log \left(\frac{20}{2}ight)$$-16500 = -2.303 	imes 2.5 	imes 8.314 	imes T 	imes \log(10)$Since $\log(10) = 1$, we have: $-16500 = -2.303 	imes 2.5 	imes 8.314 	imes T$$-16500 = -47.87 	imes T$$T = \frac{16500}{47.87} \approx 344.68 \ K$Rounding to the nearest value, $T \approx 345 \ K$.

Similar Questions

$34.05 \, mL$ of phosphorus vapour weighs $0.0625 \, g$ at $546^{\circ} C$ and $0.1 \, bar$ pressure. What is the molar mass of phosphorus?

What is the temperature of $4$ moles of a gas occupying $5 \ L$ volume at $3.32 \ bar$ (in $K$)? $\left(R=0.083 \ bar \ L \ K^{-1} \ mol^{-1}\right)$

The minimum pressure required to compress $600 \; dm^{3}$ of a gas at $1 \; bar$ to $150 \; dm^{3}$ at $40^{\circ} C$ is ......... $bar$.

An open flask at $27^{\circ}C$ contains air. To what temperature $(^{\circ}C)$ should it be heated so that one-third of the air is expelled from the flask?

The density of oxygen gas at $1.0 \, atm$ pressure and $27 \, ^oC$ temperature is ............. $g/L$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module