# An insulated piston?cylinder device initially contains 0.3 m3 of carbon dioxide at 200 kPa and...

## Question:

An insulated piston?cylinder device initially contains 0.3 {eq}m^3 {/eq} of carbon dioxide at 200 kPa and 27{eq}^{\circ} {/eq}C. An electric switch is turned on, and a 110-V source supplies current to a resistance heater inside the cylinder for a period of 10 min. The pressure is held constant during the process, while the volume is doubled. Determine the current that passes through the resistance heater.

## Current:

The current is the electric term which means when the electric charge is passed through per unit time then it will give us currently. The current is generally indicated by the letter I and it is measured in ampere. The electric current flows opposite to the direction of the flow of electron.

## Answer and Explanation:

We have given that,

Initial volume {eq}\left( {V_1 } \right) = 0.3m^3 {/eq}

Final volume {eq}\left( {V_2 } \right) = 2V_1 {/eq}

Initial pressure {eq}\left( {P_1 } \right) = 200{\rm{KPa}} {/eq}

Final pressure {eq}\left( {P_2 } \right) = 200{\rm{KPa}} {/eq}

Voltage {eq}\left( V \right) = 110V {/eq}

Time {eq}\left( {\Delta t} \right) = 10\;\min {/eq}

Initial temperature {eq}\left( {T_1 } \right) = 27^0 C = 300K {/eq}

From the properties tables of {eq}\left( {{\rm{CO}}_2 } \right) {/eq} is

Gas constant {eq}\left( R \right) = 0.1889\;{\rm{KJ/Kg}}{\rm{.K}} {/eq}

Specific heat {eq}\left( {C_P } \right) = 0.978\;{\rm{KJ/Kg}}{\rm{.K}} {/eq}

Compute the final temperature {eq}\left( {T_2 } \right) {/eq}

{eq}\begin{align*} \dfrac{{P_1 V_1 }}{{T_1 }} &= \dfrac{{P_2 V_2 }}{{T_2 }} \\ \dfrac{{200V_1 }}{{300}} &= \dfrac{{200\left( {2V_1 } \right)}}{{T_2 }} \\ T_2 &= 600K \\ \end{align*} {/eq}

Calculate the mass {eq}\left( m \right) {/eq}

{eq}m = \dfrac{{PV}}{{RT}} = \dfrac{{200 \times 0.3}}{{\left( {0.1889} \right)\left( {300} \right)}} = 1.058{\rm{Kg}} {/eq}

Calculate the work done {eq}\left( W \right) {/eq}

{eq}\begin{align*} W &= mC_P \left( {T_2 - T_1 } \right) \\ &= 1.058 \times 0.978 \times \left( {600 - 300} \right) \\ &= 310.42\;{\rm{KJ}} \\ \end{align*} {/eq}

Calculate the electric current {eq}\left( I \right) {/eq}

{eq}I = \dfrac{W}{{V\Delta t}} = \dfrac{{310.42 \times 10^3 }}{{110 \times 10 \times 60}} = 4.703\;{\rm{A}} {/eq}

Hence, this is our required solution.

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 6 / Lesson 7