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The relation of internal energy is(a) du = \(\left(\cfrac{k}{\beta}c_v\right)\) dp + \(\left(\cfrac{cp}{v\beta}-p\right)\)dv(b) du = \(\left(\cfrac{k}{\beta}c_v\right)\) dp + \(\left(\cfrac{cp}{v\beta}+p\right)\)dv(c) du = \(\left(\cfrac{k}{\beta}c_v\right)\) dp + \(\left(\cfrac{cp}{v\beta}-v\right)\)dv(a) du = \(\left(\cfrac{k}{\beta}c_v\right)\) dp + \(\left(\cfrac{cv}{v\beta}-p\right)\)dv |
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Answer» (a) du = \(\left(\cfrac{k}{\beta}c_v\right)\) dp + \(\left(\cfrac{cp}{v\beta}-p\right)\)dv |
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