A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be?

Question

1958 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-21T04:49:50

$T_{2} = 157.89 K$ $T_2 = 157.89 K$

$T_{2} = - {115.10}^{0} C$ $T_2 = -115.10 ^0C$

By changing the physical conditions like pressure, temperature and volume of the gas, the moles of the gas remain constant.
$n_{1} = n_{2}$ $n_1=n_2$

According to the ideal gas equation,

$\frac{P_{1} V_{1}}{R T_{1}} = \frac{P_{2} V_{2}}{R T_{2}}$ $(P_1V_1)/(RT_1) = (P_2V_2)/(RT_2)$

The volume of the process is also constant . As, according to the Avogadro's law,
$V α n$ $V alpha n$

So, $V_{1} = V_{2}$ $V_1=V_2$

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $(P_1)/(T_1) = (P_2)/(T_2)$

$\frac{3.8}{500} = \frac{1.2}{T_{2}}$ $3.8/500 = 1.2/ T_2$

$T_{2} = \frac{1.2}{3.8} \times 500$ $T_2 =1.2/3.8 xx500$

$T_{2} = 157.89 K$ $T_2 = 157.89 K$

$T_{2} = - {115.10}^{0} C$ $T_2 = -115.10 ^0C$

So, the new temperature will be $2280 K$ $2280 K$ .