Question

A steel gas tank of volume 0.0700 m3 is filled to the top with gasoline at 20.0 °C. The tank is placed inside a chamber with an interior temperature of 50.0 °C. The coefficient of volume expansion for gasoline is 9.50 × 10–4/C°; and the coefficient of linear expansion of steel is 12.0 × 10–6 /C°. After the tank and its contents reach thermal equilibrium with the interior of the chamber, how much gasoline has spilled? (a) 2.52 × 10–5 m 3 (b) 1.69 × 10–3 m 3 (c) 2.00 × 10–3 m 3 (d) 7.56 × 10–5 m 3 (e) 1.92 × 10–3 m 3​

Answer 1

Answer:

option (d) is your answer

Answer 2

Find the spilled volume of gasoline for given temperature changes and expansion coefficients

Explanation:

• Since the tank was filled to top by gasoline the initial volume of gas is $V_{g1}=0.07\ m^3$  and coefficient of volume expansion is $\alpha_g= 9.5x10^{-4}$
• Now let the dimension of the tank be $l, b, h$ respectively than we have ,  $V_{t1}=0.07\ m^3=lbh$  and coefficient of linear expansion $\alpha_t=12x10^{-6}$  
• Since the temperature rises from $20\ to\ 50$ °C we have, $\Delta T=50-20=30$      
• now for volume expansion of tank let let the change in volume be $\Delta V_t$then we have,                                                                                                        $V_t+\Delta V_t=(l+\Delta l)(b+\Delta b)(h+\Delta h)\\V_t+\Delta V_t=(l+l\alpha_t\Delta T )(b+ b\alpha_t\Delta T)(h+ h\alpha_t\Delta T)\\V_t+\Delta V_t=lbh(1+\alpha_t\Delta T)^3\\V_t+\Delta V_t=lbh(1+3\alpha_t\Delta T) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (since, \ \alpha_t\Delta T<<1) \\\Delta V_t=3V_t\alpha_t\Delta T\\\Delta V_t=3(0.07)(12x10^{-6})(30) \\-> \Delta V_t=0.0756x10^{-3}$   -----(a)
• now for volume expansion of gasoline we have,                                               $\Delta V_g=V_g\alpha_g\Delta T\\\Delta V_g=0.07(9.5x10^{-4})(30)\\\Delta V_g=1.995x10^{-3}$   -------(b)
• now during expansion as the gas expands the the tank is also expanding hence the volume gas of gas spilled is the difference of changes in the volume gas and tank ,          
• from (a) and (b) we get,                              $V_{spilled}= \Delta V_g-\Delta V_t\\V_{spilled}=(1.995-0.0756)x10^{-3}\\V_{spilled}= 1.92x10^{-3}\ m^3(approx.)$         ---------ANSWER