An ideal gas expands isothermally from 10^ -3 ~m^3 to 10^ -2 ~m^3 at 300 ~K against a constant pressure of 10^5 Nm^ -2 . The work done on the gas is

Correct answer: B. -900 ~J For an isothermal irreversible expansion, Work done (W)=-p_ ext (V_2- V_1) where, V_1= initial volume V_2= final volume Given, p_ ext = 10^5 Nm^ -2 , V_1=10^ -3 m^3, V_2=10^ -2 m^3 On substituting the given values in Eq, (i), We get, aligned W & =-10^5 Nm^ -2 (10^ -2 m^3-10^ -3 m^3) & =-10^5 Nm^ -2 ×10^ -3 (10-1) m^3 & =-900 Nm=-900 J aligned

NEET 2019

Exam: NEET
Year: 2019
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2019 · Chemistry · Thermodynamics (C)

An ideal gas expands isothermally from 10^ -3 ~m^3 to 10^ -2 ~m^3...

An ideal gas expands isothermally from

10^{-3} \mathrm{~m}^3

10^{-2} \mathrm{~m}^3

300 \mathrm{~K}

against a constant pressure of

10^5 \mathrm{Nm}^{-2}

. The work done on the gas is

Options

A $+270 \mathrm{~kJ}$
B $-900 \mathrm{~J}$
C $+900 k J$
D $-900 k J$

Correct answer

B. $-900 \mathrm{~J}$

Step-by-step solution

For an isothermal irreversible expansion, Work done

(\mathrm{W})=-\mathrm{p}_{\mathrm{ext}}

\left(\mathrm{V}_2- \mathrm{V}_1\right)

where,

\mathrm{V}_1=

initial volume

\mathrm{V}_2=\text{ final volume }

Given,

\mathrm{p}_{\mathrm{ext}

}=

10^5 \mathrm{Nm}^{-2}, \mathrm{ V}_1=10^{-3} \mathrm{ m}^3, \mathrm{ V}_2=10^{-2} \mathrm{ m}^3 On substituting the given values in Eq, (i), We get,

\begin{aligned} W & =-10^5 \mathrm{Nm}^{-2}\left(10^{-2} \mathrm{ m}^3-10^{-3} \mathrm{ m}^3\right) \\ & =-10^5 \mathrm{Nm}^{-2} \times10^{-3}(10-1) \mathrm{m}^3 \\ & =-900 \mathrm{Nm}=-900 \mathrm{ J} \end{aligned}

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