Answer in Mechanics | Relativity for Martin Navarro #140247
(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”
