Question

An ideal gas is brought through an isothermal compression process. The 3.00 mol of gas goes from an initial volume of 261.6×10−6 m3 to a final volume of 138.2×10−6 m3 . If 9340 J is released by the gas during this process, what are the temperature ???? and the final pressure ???????? of the gas?

Answer 1

Answer : The temperature and the final pressure of the gas is, 586.83 K and $1.046\times 10^{9}atm$ respectively.

Explanation : Given,

Initial volume of gas = $261.6\times 10^{-6}m^3$

Final volume of the gas = $138.2\times 10^{-6}m^3$

Heat released = -9340 J

First we have to calculate the temperature of the gas.

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

As per first law of thermodynamic,

$\Delta U=q+w$

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.

$q=-w$

Thus, w = -q = 9340 J

The expression used for work done will be,

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

where,

w = work done = 9340 J

n = number of moles of gas  = 3 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = ?

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

Now put all the given values in the above formula, we get the temperature of the gas.

$9340J=3mole\times 8.314J/moleK\times T\times \ln (\frac{261.6\times 10^{-6}m^3}{138.2\times 10^{-6}m^3})$

$T=586.83K$

Now we have to calculate the final pressure of the gas by using ideal gas equation.

$PV=nRT$

where,

P = final pressure of gas = ?

V = final volume of gas = $138.2\times 10^{-6}m^3=138.2\times 10^{-9}L$

T = temperature of gas = 586.83 K

n = number of moles of gas = 3 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get:

$P\times (138.2\times 10^{-9}L)=3mole\times (0.0821L.atm/mole.K)\times (586.83K)$

$P=1.046\times 10^{9}atm$

Therefore, the temperature and the final pressure of the gas is, 586.83 K and $1.046\times 10^{9}atm$ respectively.