5.0 g of nitrogen gas at 20°C and an initial pressure of

Step 1 of 5

a.) We need to find out the gas volume and temperature after the expansion.

From the ideal gas equation,

\(p_{1} V_{1}=n R T_{1}\)

Implies,

\(V_{1}=\frac{n R T_{1}}{p_{1}}\)

We know that,

\(n =\frac{M}{M_{m o l}}\)

\(n =\frac{5 g}{28 g / m o l}=0.1786 \mathrm{~mol}\)

The initial pressure is, \(p_{1}=3 a t m=304000 \mathrm{~Pa}\)

The initial temperature is, \(T_{1}=20^{0} \mathrm{C}=293 \mathrm{~K}\)

So,

\(V_{1}=\frac{(0.1786 \mathrm{~mol})(8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K})(293 \mathrm{~K})}{304000 \mathrm{~Pa}}\)

The process is isobaric expansion that is \(p_{1}=p_{2}\)

So we can write,

\(\frac{V_{2}}{T_{2}}=\frac{V_{1}}{T_{1}}\)

The volume has tripled during the process.

That is \(V_{2}=3 V_{1}\)

So,

\(T_{2} =\frac{3 V_{1} T_{1}}{V_{1}}\)

\(T_{2} =3 T_{1}=3 \times 273 \mathrm{~K}\)

\(T_{2} =879 \mathrm{~K}\)