Question

A volume of gas starts at a pressure of 10 atmospheres (atm) and a temperature of 27°C. If the temperature is increased by 75°C and the volume of the gas remains constant, what is the new pressure?

Answer 1

Given that,

Initial pressure = 10 atm

Initial temperature = 27 °C

Final temperature = 75 °C

We have to find final pressure of the gas$.$

❖ As per Gay Lussac's law at constant volume, pressure of an ideal gas is directly proportional to the temperature.

Mathematically, P ∝ T

First of all we have to convert the temperature into kelvin.

⟩ Initial temperature = 27 °C = 300 K

⟩ Final temperature = 75 °C = 348 K

By substituting the given values;

$\sf:\implies\:\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}$

$\sf:\implies\:\dfrac{10}{P_2}=\dfrac{300}{348}$

$\sf:\implies\:P_2=\dfrac{10\times 348}{300}$

$\sf:\implies\:P_2=\dfrac{348}{30}$

$:\implies\:\underline{\boxed{\bf{\gray{P_2=11.6\:atm}}}}$

Answer 2

Answer:

$\green{ \boxed {\sf \: Final \: pressure \: of \: the \: gas \: \: P_2 = 11.6 \: atm}}$

Given,

Initial pressure $P_1$ = 10 atm

Initial temperature $T_1$ = 27 °C = 27+273=300 K

Final temperature $T_2$ = 75 °C =75+273= 348 K

Final pressure of the gas$P_2$ = to find

☞ Here volume is constant.

So we know that,

$\: \: \: \sf \frac{P_1}{T_1} = \frac{P_2}{T_2}$

Now by substituting the given values;

$\: \: \: \: \: \: \sf \: \frac{P_1}{T_1} = \frac{P_2}{T_2} \\ \sf \implies \: \frac{10}{300} = \frac{P_2}{348} \\ \sf \implies \: P_2 = \frac{10 \times 348}{300} \\ \sf \implies \: P_2 = 11.6 \: \: atm$

So,

$\green{ \boxed {\sf \: Final \: pressure \: of \: the \: gas \: \: P_2 = 11.6 \: atm}}$