(D) According to the first law of thermodynamics,$\Delta U = q + w$. Given: Heat absorbed $q = +100 \ J$. Work done on the gas $w = -P_{ext} \Delta V$

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(D) According to the first law of thermodynamics,$\Delta U = q + w$. Given: Heat absorbed $q = +100 \ J$. Work done on the gas $w = -P_{ext} \Delta V$. $P_{ext} = 1.5 \ atm$,$\Delta V = V_f - V_i = 2.0 \ L - 8.0 \ L = -6.0 \ L$. $w = -(1.5 \ atm) \times (-6.0 \ L) = +9.0 \ L \cdot atm$. Convert work to Joules: $w = 9.0 \times 101.32 \ J = 911.88 \ J$. Therefore,$\Delta U = 100 \ J + 911.88 \ J = 1011.88 \ J \approx 1011.9 \ J$.

Class 11 Chemistry Thermodynamics First law of thermodynamics

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$A$ gas absorbs $100 \ J$ of heat and is simultaneously compressed by a constant external pressure of $1.5 \ atm$ from a volume of $8.0 \ L$ to $2.0 \ L$. The change in internal energy for the gas in Joules is $(1 \ L \cdot atm = 101.32 \ J)$.

AP EAMCET 2017 All AP EAMCET

A
$-1011.9$
B
$-909.9$
C
$+909.9$
D
$1011.9$

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$A$ gas absorbs $100 \ J$ of heat and is simultaneously compressed by a constant external pressure of $1.5 \ atm$ from a volume of $8.0 \ L$ to $2.0 \ L$. The change in internal energy for the gas in Joules is $(1 \ L \cdot atm = 101.32 \ J)$.

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