Is the formula of work done in thermodynamics dW= - (external pressure)×dV also valid for liquids and solids???
Answers
No , it's not valid for solids and liquids.
Answer:
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Explanation:
=w=−P
external
×ΔVstart text, w, o, r, k, end text, equals, start text, w, end text, equals, minus, start text, P, end text, start subscript, start text, e, x, t, e, r, n, a, l, end text, end subscript, times, delta, start text, V, end text
where \text P_{\text{external}}P
external
start text, P, end text, start subscript, start text, e, x, t, e, r, n, a, l, end text, end subscript is the external pressure (as opposed to the pressure of the gas in the system) and \Delta \text VΔVdelta, start text, V, end text is the change in the volume of the gas, which can be calculated from the initial and final volume of the gas:
\Delta \text V=\text V_{\text {final}}-\text V_{\text {initial}}ΔV=V
final
−V
initial
delta, start text, V, end text, equals, start text, V, end text, start subscript, start text, f, i, n, a, l, end text, end subscript, minus, start text, V, end text, start subscript, start text, i, n, i, t, i, a, l, end text, end subscript
Since work is energy, it has units of Joules (where 1\,\text J= 1\,\dfrac {\text {kg} \cdot \text m^2}{\text s^2}1J=1
s
2
kg⋅m
2
1, start text, J, end text, equals, 1, start fraction, start text, k, g, end text, dot, start text, m, end text, squared, divided by, start text, s, end text, squared, end fraction). You may also see other units used, such as atmospheres for pressure and liters for volume, resulting in \text L \cdot \text{atm}L⋅atmstart text, L, end text, dot, start text, a, t, m, end text as the unit for work. We can convert \text L \cdot \text{atm}L⋅atmstart text, L, end text, dot, start text, a, t, m, end text to convert to Joules using the conversion factor of \dfrac{101.325\,\text J}{1\,\text L\cdot \text{atm}}
1L⋅atm
101.325J
start fraction, 101, point, 325, start text, J, end text, divided by, 1, start text, L, end text, dot, start text, a, t, m, end text, end fraction.