Question

A rigid container holds $0.20g$of hydrogen gas. How much heat is needed to change the temperature of the gas

a.From$50K$ torole="math" localid="1648534484983" $100K?$

b.From localid="1648534491176" $250K$to localid="1648534494324" $300K?$

c.From localid="1648534497013" $2250K$to localid="1648534500972" $2300K?$

Short answer

The heat change in temperature of gas from,

$\left(a\right)$$50K$to $100K$is $62J.$

$\left(b\right)$$250K$to $300K$is $104\mathrm{J}.$

$\left(c\right)$F$2250K$to $3000K$is $150\mathrm{J}.$

Step-by-step solution

Step: 1 a Temperature changes from 50K to 100K:

If the temperature changes by $∆T$,then the thermal energy for diatomic gas changes by equation $\left(20.30\right)$in the form

$\mathrm{\Delta }{E}_{\mathrm{th}}=n{C}_V\mathrm{\Delta }T$

Knowing the mass $M$and the molar mass $m$, we can get the number of moles by

$n=\frac{M}{m}=\frac{0.2\mathrm{g}}{2\mathrm{g}/\mathrm{mol}}=0.1\mathrm{mol}$

This thermal energy converts to heat..The temperature change from$50K$to $100K$. In this range,$C_V=\frac{3}{2}R$. So, the heat in equation will be

$\begin{array}{r}\mathrm{\Delta }{E}_{\mathrm{th}}=n{C}_V\mathrm{\Delta }T\\ =\frac{3}{2}nR\mathrm{\Delta }T\\ =\frac{3}{2}\left(0.1\mathrm{mol}\right)(8.314\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})(100\mathrm{K}-50\mathrm{K})\\ =62\mathrm{J}.\end{array}$

Step: 2 b Temperature changes from 250K to 300K:

The temperature changes from $250K$to $300K$.In this range,$C_V=\frac{5}{2}R$.so,the heat equation will be

$\begin{array}{r}\mathrm{\Delta }{E}_{\mathrm{th}}=n{C}_V\mathrm{\Delta }T\\ =\frac{5}{2}nR\mathrm{\Delta }T\\ =\frac{5}{2}\left(0.1\mathrm{mol}\right)(8.314\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})(300\mathrm{K}-250\mathrm{K})\\ =104\mathrm{J}.\end{array}$

Step: 3 c Temperature changes from 2250K to 3000K:

The temperature changes from $2250K$to $3000K$.In this range,$C_V=\frac{7}{2}R$.so,the heat equation will be

$\begin{array}{r}\mathrm{\Delta }{E}_{\mathrm{th}}=n{C}_V\mathrm{\Delta }T\\ =\frac{7}{2}nR\mathrm{\Delta }T\\ =\frac{7}{2}\left(0.1\mathrm{mol}\right)(8.314\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})(2300\mathrm{K}-2250\mathrm{K})\\ =150\mathrm{J}.\end{array}$