Question

Three moles of neon expand isothermally from 0.163 m3 to 0.254 m3 while 5.09 × 103 J of heat flows into the gas. Assuming that neon is an ideal gas, find its temperature.

Answer 1

Answer :  The temperature is, 500.78 K

Explanation :

According to the question, this is the case of isothermal reversible expansion of gas.

As per first law of thermodynamic,

$q=\Delta U+w$

where,

$\Delta U$ = internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

$\Delta U=0$

$q=w$

The expression used for work done will be,

$w=nRT\ln (\frac{V_2}{V_1})$

where,

w = work done on the system = $5.09\times 10^3J$

n = number of moles of gas  = 3 moles

R = gas constant = 8.314 J/mole K

T = temperature of gas  = ?

$V_1$ = initial volume of gas  = $0.163m^3$

$V_2$ = final volume of gas  = $0.245m^3$

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

$5.09\times 10^3J=-3mole\times 8.314J/moleK\times T\times \ln (\frac{0.245m^3}{0.163m^3})$

$T=500.78K$

Thus, the temperature is, 500.78 K