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3 years ago

8

If 3.61 m3 of a gas initially at STP is placed under a pressure of 2.67 atm , the temperature of the gas rises to 37.9 ∘C. What

is the volume?

1 answer:

Pavel [41]3 years ago

7 0

Answer: $1.54m^3$

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's 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 at STP = 1 atm

$P_2$ = final pressure of gas = 2.67 atm

$V_1$ = initial volume of gas = $3.61m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas at STP = $0^oC=273+0=273K$

$T_2$ = final temperature of gas = $37.9^oC=273+37.9=310.9K$

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

$\frac{1atm\times 3.61m^3}{273K}=\frac{2.67\times V_2}{310.9K}$

$V_2=1.54m^3$

Thus the final volume will be $1.54m^3$

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