One mole of an ideal gas expands isothermally | vedclass

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

Question

(B) For an isothermal process,$\Delta T = 0$,therefore the change in internal energy $\Delta U = 0$. For a reversible isothermal expansion,the work do

Vedclass · Accepted Answer

$0, 1.718 \ kJ, -1.718 \ kJ$

Vedclass · Answer

(B) For an isothermal process,$\Delta T = 0$,therefore the change in internal energy $\Delta U = 0$. For a reversible isothermal expansion,the work done $w = -nRT \ln(V_2/V_1)$. Given $n = 1 \ mol$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$V_1 = 10 \ dm^3$,$V_2 = 20 \ dm^3$. $w = -1 	imes 8.3 	imes 300 	imes \ln(20/10) = -2490 	imes \ln(2)$. Using $\ln(2) = 2.303 	imes \log(2) = 2.303 	imes 0.30 = 0.6909$. $w = -2490 	imes 0.6909 = -1720.34 \ J \approx -1.718 \ kJ$. Since $\Delta U = q + w = 0$,we have $q = -w = 1.718 \ kJ$. Thus,$\Delta U = 0, q = 1.718 \ kJ, w = -1.718 \ kJ$.

One mole of an ideal gas expands isothermally and reversibly from $10 \ dm^3$ to $20 \ dm^3$ at $300 \ K$. $\Delta U$,$q$ and work done in the process respectively are $:$ Given $: R=8.3 \ J \ K^{-1} \ mol^{-1}$,$\ln 10=2.3$,$\log 2=0.30$,$\log 3=0.48$

Similar Questions

If the heat of combustion of isobutane at constant volume is $Q_v$,what will be the heat of combustion of one mole of isobutane at constant pressure and standard conditions? (Given: $\Delta n_g = -3.5$)

The temperature of a system decreases in which of the following processes?

If $Q$ is the heat liberated from the system and $W$ is the work done on the system,then the first law of thermodynamics can be written as:

$A$ sample of gas expands against a constant external pressure of $1 \ atm$ from a volume of $2 \ L$ to $12 \ L$. During this process,it absorbs $600 \ J$ of heat. Calculate $\Delta E$ of the system.

Vedclass Test Series

Exam Paper Generator

Online Exam Module