A beaker with a metal bottom is filled with $20\mathrm{g}$of water at $20°\mathrm{C}$. It is brought into good thermal contact with a $4000{\mathrm{cm}}^3$container holding $0.40\mathrm{mol}$of a monatomic gas at $10atm$pressure. Both containers are well insulated from their surroundings.

What is the gas pressure after a long time has elapsed? You can assume that the containers themselves are nearly massless and do not affect the outcome.

Amount of water in beaker$=20g$

Temperature $={20}^{\circ }C$

Weight of container role="math" localid="1648190396023" $=4000c{m}^3$

Amount of monatomic gas $=0.40mol$

Pressure$=10atm$

$\Delta Q=mc\Delta T$

$=mc\left(T_f-T_i\right)$

Here,

$m$is the mass of the substance, $c$is the specific heat capacity of the substance,

$T_i$the initial temperature of the substance, and is the final temperature of the substance.

Based on the ideal gas equation,

$p_{\mathrm{gas}}{V}_{\mathrm{gas}}=n_{\mathrm{gas}}R{T}_{\text{igas}}$

Modify the terms of $T_{\text{gas}}$

$T_{\text{igas}}=\frac{p_{\text{gas}}{V}_{\text{gas}}}{n_{\text{gas}}R}$

Substitute the values of

$10atm$for $P_{gas}$

localid="1648191465577" $4000c{m}^3$for $V_{gas}$

$0.40mol$for $n_{gas}$

And $8.314\mathrm{J}·{\mathrm{k}}^{-1}·{\mathrm{mol}}^{-1}$for $R$in the equation.

$T_{\text{igas}}=\frac{10\mathrm{atm}\left(\frac{1.013\times {10}^5\mathrm{pa}}{1\mathrm{atm}}\right)\left(4000{\mathrm{cm}}^3\left(\frac{{10}^{-6}{\mathrm{m}}^3}{1{\mathrm{cm}}^3}\right)\right)}{0.40\mathrm{m}\left(8.314\mathrm{J}·{\mathrm{k}}^{-1}·{\mathrm{mol}}^{-1}\right)}$

$=\frac{10\left(1.013\times {10}^5\mathrm{pa}\right)\left(4000\times {10}^{-6}{\mathrm{m}}^3\right)}{0.40\mathrm{m}\left(8.314\mathrm{J}·{\mathrm{k}}^{-1}·{\mathrm{mol}}^{-1}\right)}$

$=1219\mathrm{K}$$=1219\backslash mathrm\{~K\}\backslash end\left\{aligned\right\}\$$

A water system and a gas system interact here. The equilibrium of a closed system depends on the transfer of heat between interacting systems.

$Q_{\text{air}}+Q_{\text{water}}=0\mathrm{J}$

$n_{gas}{C}_V\left(T_f-T_{\text{igas}}\right)+m_{\text{water}}{c}_{\text{water}}\left(T_f-T_{\text{fwater}}\right)=0\mathrm{J}$

Here,

Specific heat at constant volume$=C_V$

Final temperature of both the gas and water. $=T_f$

Convert the unit of mass from $\mathrm{g}$to $\mathrm{kg}$

$m_w=20\mathrm{g}\left(\frac{{10}^{-3}\mathrm{kg}}{\lg }\right)$

$=20\times {10}^{-3}\mathrm{kg}$

Convert the unit of temperature from ${}^0\mathrm{C}$to $\mathrm{K}$

$T_{\text{iwater}}=\left(20{}^{\circ }\mathrm{C}+273\right)\mathrm{K}$

$=293\mathrm{K}$

Modify the equation (1) in terms of $T_f$

$T_f=\frac{m_{\text{water}}{c}_{\text{water}}{T}_{\text{iwater}}+n_{gas}{C}_V{T}_{\text{igas}}}{n_{\text{gas}}{C}_V+m_{\text{water}}{c}_{\text{water}}}$

Substitute the values of:

$20\times {10}^{-3}\mathrm{kg}$for $m_{water},$,

$12.471J/molK$for $C_V$,

$4186J/kgK$for $c_{water}$,

$293K$for $T_{iwater}$,

$1219K$for $T_{\text{igas}}$,

And $0.40\mathrm{mol}$for $n_{\text{gas}}$$T_f=\frac{m_{\text{water}}{c}_{\text{water}}{T}_{\text{moter}}+n_{\text{gas}}{C}_V{T}_{\text{igas}}}{n_{\text{gas}}{C}_V+m_{\text{water}}{c}_{\text{water}}}$

localid="1648193216246" $=\frac{\left(20\times {10}^{-3}\mathrm{kg}\right)(4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K})\left(293\mathrm{K}\right)+\left(0.40\mathrm{mol}\right)(12.471\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})\left(1219\mathrm{K}\right)}{\left(0.40\mathrm{mol}\right)(12.471\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})+\left(20\times {10}^{-3}\mathrm{kg}\right)(4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K})}$

$=345.04\mathrm{K}$

Calculate the final pressure of the gas by using the ideal gas equation.

$p_{gas}^{\text{'}}=\frac{n_{gas}R{T}_f}{V_{gas}}$

Substitute $0.40\mathrm{mol}$for $n_{gas}$,

$8.314{\mathrm{J}}^{-1}{\mathrm{k}}^{-1}·{\mathrm{mol}}^{-1}=R$

role="math" localid="1648193448546" $345.04K=T_f$

And $4000{\mathrm{cm}}^3=V_{gas}$

$p_{gas}^{\text{'}}=\frac{(0.40\mathrm{mol})\left(8.314\mathrm{J}·{\mathrm{k}}^{-1}·{\mathrm{mol}}^{-1}\right)(345.19\mathrm{K})}{4000{\mathrm{cm}}^3\left(\frac{{10}^{-6}{\mathrm{m}}^3}{1{\mathrm{cm}}^3}\right)}$

$=2.83atm$

Hence, the gas pressure after long time elapsed is$2.83atm.$