Chapter 10, Problem 10.62QA

Interpretation Introduction

To find:

The effect of the following conditions on the pressure of gas:

a) The absolute temperature is halved, and the volume is doubled.

b) Both the absolute temperature and the volume are doubled.

c) The absolute temperature is increased by $75 \%,$ and the volume is decreased by $50 \%$.

Answer to Problem 10.62QA

Solution:

a) When the absolute temperature is halved and the volume is doubled, the pressure would be ${(1/4)}^{th}$ of the initial pressure

b) When both the absolute temperature and the volume are doubled, there would be no change in the pressure.

c) When the absolute temperature is increased by $75\mathrm{ }\mathrm{\%}$ and the volume is decreased by $50\mathrm{ }\mathrm{\%},$ the pressure would increase $3.5$ times the initial pressure.

Explanation of Solution

1) Concept:

To find the change in the pressure of a gas, we have to apply the combined gas law for the given three conditions. We have to plug the values for final temperatures and volumes as described in the condition to get a value of the final pressure.

2)  Formula:

The combined gas law

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

3)  Given:

i) Absolute temperature $T_2=\frac{1}{2} T_1$ and volume $V_2=2 V_1$

ii) Absolute temperature $T_2=2 T_1$ and volume $V_2=2 V_1$

iii) $75 \%$ increase in absolute temperature and $50 \%$ decrease in volume.

4)  Calculations:

a) The absolute temperature is halved and the volume doubled.

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Putting $T_2=\frac{1}{2} T_1$ and $V_2=2 V_1$

$\frac{P_1{V}_1}{T_1}=\frac{P_{2}2 V_1}{\frac{1}{2} T_1}$

$P_2=\frac{1}{4}{P}_1$

When the absolute temperature is halved and the volume doubled, the pressure would be ${(1/4)}^{th}$ of the initial pressure.

b) Both the absolute temperature and the volume doubled.

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Putting $T_2=2 T_1$ and $V_2=2 V_1$

$\frac{P_1{V}_1}{T_1}=\frac{P_{2}2 V_1}{2 T_1}$

$P_2=P_1$

If both the absolute temperature and the volume are doubled, there would be no change in pressure.

c) The absolute temperature increased by $75 \%$ and the volume decreased by $50 \%$.

$T_2=T_1+0.75 T_1=1.75 T_1$

$V_2=V_1-0.50 V_1=0.50 V_1$

Applying the combined gas law

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Inserting $T_2=1.75 T_1$ and $V_2=0.50 V_1$

$\frac{P_1{V}_1}{T_1}=\frac{P_2\times 0.50 V_1}{1.75 T_1}$

$P_2=\frac{1.75}{0.50} P_1$

$P_2=3.5 P_1$

When the absolute temperature is increased by $75\mathrm{ }\mathrm{\%}$ and the volume is decreased by $50\mathrm{ }\mathrm{\%}$, the pressure would increase $3.5$ times the initial pressure.

Conclusion:

Applying the combined gas law, we have estimated the change in pressure for all the three conditions.