(A) Given: $n = 2 \ mol$,$V_1 = 10 \ dm^3 = 10 \times 10^{-3} \ m^3 = 0.01 \ m^3$,$V_2 = 2 \ m^3$,$P_{ext} = 101.325 \ kPa = 101.325 \times 10^3 \ Pa$

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(A) Given: $n = 2 \ mol$,$V_1 = 10 \ dm^3 = 10 \times 10^{-3} \ m^3 = 0.01 \ m^3$,$V_2 = 2 \ m^3$,$P_{ext} = 101.325 \ kPa = 101.325 \times 10^3 \ Pa$.The formula for work done in an irreversible expansion is $W = -P_{ext} \cdot \Delta V$.$\Delta V = V_2 - V_1 = 2 \ m^3 - 0.01 \ m^3 = 1.99 \ m^3$.$W = -101.325 \times 10^3 \ Pa \times 1.99 \ m^3$.$W = -201636.75 \ J = -201.63675 \ kJ$.Rounding to one decimal place,$W = -201.6 \ kJ$.

Class 11 Chemistry Thermodynamics First law of thermodynamics

Difficult

Two moles of an ideal gas are allowed to expand from a volume of $10 \ dm^3$ to $2 \ m^3$ at $300 \ K$ against a pressure of $101.325 \ kPa$. Calculate the work done. (in $kJ$)

MHT CET 2018 All MHT CET

A
$-201.6$
B
$13.22$
C
$-810.6$
D
$-18.96$

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Two moles of an ideal gas are allowed to expand from a volume of $10 \ dm^3$ to $2 \ m^3$ at $300 \ K$ against a pressure of $101.325 \ kPa$. Calculate the work done. (in $kJ$)

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