Question #320958

    A gas in a cylinder expands from a volume of 0.120 m3 to 0.320 m3. Heat flows into the gas just rapidly enough to keep the pressure constant at 1.60 x 105 Pa during the expansion. The total heat added is 1.15 x 105 J. (a) Find the work done by the gas. (b) Find the change in internal energy of the gas. 




1
Expert's answer
2022-04-01T08:43:06-0400

WorkdonebythegasduringexpansionatconstantpressureWork \>done \>by\> the\>gas\>during\>expansion\>at\>constant\>pressure

W=pdV∆W = -pdV

W=pd(V2V1)∆W = -pd(V_{2}-V_{1})

W=1.60×105pa(0.320m30.120m3)∆W = -1.60\times10^5pa(0.320m^3-0.120m^3)

=0.32×105pa.m3=-0.32\times10^5pa.m^3

=32000pa.m3= -32000 pa.m^3

=32000(N/m2).(m3)= -32000(N/m^2).(m^3)

=32000N.m= -32000 N.m

W=32000Joule∆W = -32000 Joule

Workdonebythesystem=32000JouleWork\> done\> by\> the \> system=-32000 \>Joule

(b)Changeintheinternalenergyofthegas(b) Change\>in\>the\>internal\>energy\>of\>the\>gas

U=Q+W∆U = ∆Q + ∆W

GivenGiven

Q=1.15×105Joule∆Q = 1.15\times10^5 Joule

W=32000Joule∆W = -32000 Joule

U=1.15×105J32000Joule∆U = 1.15\times10^5J-32000Joule

U=83000JouleAns.∆U=83000\>Joule\>Ans.





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