Question

The temperature of an ideal gas in a sealed 0.1 m3 container is reduced from 430 K to 270 K. The final pressure of the gas is 70 kPa. The molar heat capacity at constant volume of the gas is 28.0 J/mol · K. The work done by the gas is closest to:a) -42
b) -54


c) 54


d) 42


e) 0.00

Answer 1

Answer:

Thus, the heat absorbed by the gas is closest to 13.97 kJ

Explanation:

First we have to calculate the moles of gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of gas = 70 kPa  = 70000 Pa

V = Volume of gas = 0.1m^3

n = number of moles = ?

R = Gas constant = $8.314m^3Pa/mol.K$

T = Temperature of gas = 270K

Putting values in above equation, we get:

$70000Pa\times 0.1m^3=n\times (8.314m^3Pa/mol.K)\times 270K\\n = 3.118mol$

Heat released at constant volume is known as internal energy.

The formula used for change in internal energy of the gas is:

$:\Delta Q=\Delta U\\\\\Delta U=nC_v\Delta T\\\\\Delta Q=nC_v(T_2-T_1)$

where,

$\Delta Q$ = heat at constant volume = ?  

$\Delta U$ = change in internal energy

n = number of moles of gas = 3.118 moles

$C_v$= heat capacity at constant volume gas = 28.0 J/mol.K

$T_1$= initial temperature = 430 K

$T_2$ = final temperature = 270 K

Now put all the given values in the above formula, we get:

$\Delta Q=nC_v(T_2-T_1)\Delta Q=(3.118moles)\times (28.0J/mol.K)\times (270-430)K\Delta Q=-13970.19J\\\\=-13.97kJ$

Thus, the heat absorbed by the gas is closest to 13.97 kJ