Question

A gas that has a volume of 28 liters, a temperature of 42°C, and an unknown pressure has its volume increased to 49 liters and its temperature decreased to 27°C. If the pressure is measured after the change at 4.0 atm, what was the original pressure of the gas?

Answer 1

Answer:

7.35 atm

Explanation:

From the question we are given;

• Initial volume, V1 as 28 L
• Initial temperature, T1 as 42°C or 315 K
• New volume. V2 as 49 L
• New temperature, T2 as 27°C  or 300 K
• New Pressure, P2 as 4.0 atm

We are required to determine the initial pressure, P1

We are going to use the combined gas law...

According to the combined gas law,

$\frac{P_{1} V_{1} }{T_1} =\frac{P_2V_2}{T_2}$

Therefore, to get P1

$P_1=\frac{P_2V_2T_1}{V_1T_2}$

We get;

$P_1=\frac{(4.0)((49)(315)}{(300)(28)}$

$P_1=7.35$

Therefore, the initial pressure of the gas is 7.35 atm