Question

25cm3 of gas at 1.0atm at 25℃ was compressed to 15cm3 at 35℃,calculate the final pressure​

Answer 1

PV =nRt (ideal gas equation )

V1=25cm3

T1=25°c

p1=1 atm

V2=15cm3

T2=35°c

p2=?

since number of moles of gas are constant,

p1.v1/T1= p2.v2/T2

=> 1*25/25= p2* 15/35

p2= 35/15

=>p2=2.3 atm

therefore final pressure is 2.3atm

Answer 2

Answer:

Final pressure of the gas will be 1.72 atm.

Explanation:

Given that -

• P1 = 1.0 atm
• V1 = 25 cm³
• V2 = 15 cm³
• T1 = 25°C
• T2 = 35°C

To find -

• Final pressure of gas.

Solution -

First of all, let we change the temperature into Kelvin.

T1 = 25°C

$\longrightarrow$ T1 = 25 + 273

$\longrightarrow$ T1 = 298 K

T2 = 35°C

$\longrightarrow$ T2 = 35 + 273

$\longrightarrow$ T2 = 308 K

Using combined gas law,

$\large{\boxed{\sf{\blue{\dfrac{P_{1} V_{1}}{T_1} = \dfrac{P_{2} V_{2}}{T_2}}}}}$

$\longrightarrow$ $\sf{P_{2} = \dfrac{P_{1} V_{1} T_{1}}{V_{2} T_{1}}}$

Subsituting the values,

$\longrightarrow$ $\sf{P_{2} = \dfrac{1 \times 25 \times 308}{15 \times \298}}$

$\longrightarrow$ $\sf{P_{2} = \dfrac{7700}{4470}}$

$\longrightarrow$ $\sf{P_{2} = 1.72\:atm}$

Hence, final pressure of the gas will be 1.72 atm.