Question

Calculate the final pressure, in atmospheres, for each of the following, if $V$ and$n$ do not change:

a. A gas with an initial pressure of $1.20atm$ at$75°C$ is cooled to $-32°C.$

b. A sample of $N$, with an initial pressure of $780.mmHg$ at $-75°C$ is heated to$28°C.$

Short answer

The final pressure, in atmospheres, for each of the following :

a.$0.512atm$

b.$291.2mmHg$

Step-by-step solution

Part(a) Step 1: Given information

We need to find that the final pressure, in atmospheres, if $V$and $n$ do not change .

Part(a) Step 2: Explanation

Here , we will use Gay-Lussac's law which says ;

We are given with ,$P_1=1.20atm,T_2=32°C,T_1=75°C$

$\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow P_2=\frac{P_1\times T_2}{T_1}=\frac{1.20\times \left(-32\right)}{75}=0.512atm$

Part (b) Step 1: Given information 

The final pressure, in atmospheres, if $V$and $n$do not change .

Part(b) Step 2: Explanation

By using gay-lussac's law ;

We are given that ; $P_1=780mmHg,T_2=28°C,T_1=-75°C$

Substituting these values in given formula ;

$\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow P_2=\frac{P_1\times T_2}{T_1}=\frac{780\times \left(28\right)}{\left(-75\right)}=291.2atm$