Answer to Question #140247 in Mechanics | Relativity for Martin Navarro

Question

The work required to compress a gas reversibly according to pV^1.30 = c is 67,790J, if there is no flow. Determine ΔU and Q if the gas is (a) air, (b) methane. For methane, k= 1.321, R=518.45 J/kg., Csubv= 1.6187, csubp= 2.1377 kJ/kg.K.

Answer 1

(a) The amount of heat transferred by the air is,

$Q=(\frac{K_{air} - n}{K_{air} - 1})W$

$K_{air}=$ specific heat ratio of air

$W=$ work required to compress the gas reversibly

substitute 1.4 for $K_{air}$$, 1.3$ for $n$ and $-67,790J$ for $W$

$Q= (\frac{1.4-1.30}{1.4-1})(-67,790J)=-16947.5J=-16.95kJ$

The amount of heat transferred by the air ($Q$ ) is $16.95kJ$

change in internal energy of the air is,

$Q=\Delta U+W$

$Q=-16947.5J$ and $W=-67,790J$

$-16947.5J=\Delta U-67,790$

$\Delta U=-16947.5J+67,790J$

$\Delta U = 50842.5J= 50.84kJ$

(b) Amount of heat transferred by the methane,

$Q=(\frac{K_{methane}-n}{K_{methane}-1})W$

$K_{methane}=1.321, n=1.30 ,W=-16,790J$

$Q=(\frac{1.321-1.30}{1.321-1})(-67,790)=-4434.86J=-4.43kJ$

the amount of heat transferred by the methane ($Q$ ) is $4.43kJ$

Change in internal energy of the methane,

$Q=\Delta U+W$

$-4434.86J=\Delta U-67,790J$

$\Delta U= -4434.86J+67,790J=63355.14J=63.36kJ$