Answer to Question #320958 in Molecular Physics | Thermodynamics for NICKO

Question

    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. 

Answer 1

$Work \>done \>by\> the\>gas\>during\>expansion\>at\>constant\>pressure$

$∆W = -pdV$

$∆W = -pd(V_{2}-V_{1})$

$∆W = -1.60\times10^5pa(0.320m^3-0.120m^3)$

$=-0.32\times10^5pa.m^3$

$= -32000 pa.m^3$

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

$= -32000 N.m$

$∆W = -32000 Joule$

$Work\> done\> by\> the \> system=-32000 \>Joule$

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

$∆U = ∆Q + ∆W$

$Given$

$∆Q = 1.15\times10^5 Joule$

$∆W = -32000 Joule$

$∆U = 1.15\times10^5J-32000Joule$

$∆U=83000\>Joule\>Ans.$