At the $1$ bar pressure,the volume of a gas is $0.6 \ L$. If the gas receives $122 \ J$ of heat at $1$ bar pressure,the volume becomes $2 \ L$. Calculate the change in its internal energy. $(1 \ L \cdot bar = 101.32 \ J)$
(A) Given: $q = 122 \ J$,$P = 1 \ bar$,$V_1 = 0.6 \ L$,$V_2 = 2 \ L$.
Change in volume $\Delta V = V_2 - V_1 = 2 - 0.6 = 1.4 \ L$.
Work done $w = -P \cdot \Delta V = -1 \ bar \times 1.4 \ L = -1.4 \ L \cdot bar$.
Converting work to Joules: $w = -1.4 \times 101.32 \ J = -141.848 \ J$.
According to the first law of thermodynamics: $\Delta U = q + w$.
$\Delta U = 122 \ J + (-141.848 \ J) = -19.848 \ J$.
Rounding to two decimal places,the change in internal energy is $-19.85 \ J$.
Change in volume $\Delta V = V_2 - V_1 = 2 - 0.6 = 1.4 \ L$.
Work done $w = -P \cdot \Delta V = -1 \ bar \times 1.4 \ L = -1.4 \ L \cdot bar$.
Converting work to Joules: $w = -1.4 \times 101.32 \ J = -141.848 \ J$.
According to the first law of thermodynamics: $\Delta U = q + w$.
$\Delta U = 122 \ J + (-141.848 \ J) = -19.848 \ J$.
Rounding to two decimal places,the change in internal energy is $-19.85 \ J$.
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