Question

Two moles of an ideal gas at 2atm and 27°C is compressed isothermally to one half of its volume by a constant external pressure of 4atm. Calculate q, w ∆E. If R=0.082L ATM mol^-1K^-1​

Answer 1

Answer:

q =w = 3410 J,

∆E =0 (since for isothermal process ∆T=0)

Explanation:

pls see the attachment

Answer 2

Answer : The value of q, w and U for the reversible, isothermal compression are,  -3458.32 J, 3458.32 J and 0 J respectively.

Explanation : Given,

Moles of gas = 2 mole

Initial pressure of gas = 2 atm

Final pressure of the gas = 4 atm

Temperature of the gas = $27^oC=273+27=300K$

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.

The expression used for work done will be,

$w=-2.303nRT\log (\frac{P_1}{P_2})$

where,

w = work done on the gas

n = number of moles of gas

R = gas constant = 8.314 J/mole K  or 0.0821 L.atm/mol.K

T = temperature of gas

n = moles of the gas

$P_1$ = initial pressure of gas

$P_2$ = final pressure of gas

Now put all the given values in the above formula, we get the work done.

$w=-2.303\times 2mole\times 8.314J/moleK\times 300K\times \log (\frac{2atm}{4atm})$

$w=3458.32J$

And we know that, the heat is equal to the work done with opposite sign convention.

So, $q=-3458.32J$

Therefore, the value of q, w and U for the reversible, isothermal compression are, -3458.32 J, 3458.32 J and 0 J respectively.