Question

The pressure P of a gas varies directly with the temperature T and inversely with the volume V. A certain gas has a volume of 6 liters (L), a temperature of 400 Kelvins (K), and a pressure of 2.4 atmospheres (atm). If the gas is expanded to a volume of 15 L and is heated to 650 K, what will the new pressure be?

Answer 1

Answer:

The value of new pressure will be 1.56 atm.

Explanation:

Using combined gas law :

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 2.4 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = $6 L$

$V_2$ = final volume of gas =  15 L

$T_1$ = initial temperature of gas = $400 K$

$T_2$ = final temperature of gas = $650 K$

Now put all the given values in the above equation, we get:

$\frac{2.4 atm \times 6 L}{400 K}=\frac{P_2\times 15 L}{650 K}$

$P_2=\frac{2.4 atm \times 6 L\times 650 K}{400 K\times 15 L}$

$P_2=1.56 atm$

The value of new pressure will be 1.56 atm.