Question

A car tyre has air at 1.5 atm at 300 K. If P increases to 1.75 atm at constant volume, the temperature will be​

Answer 1

Given:

Initial pressure of air in tyre,P₁= 1.5 atm

Final pressure of air in tyre,P₂= 1.75 atm

Initial temperature of air in tyre,T₁= 300 K

To Find:

Final temperature of air

Solution:

• According to equation of Ideal gas,

$\pink{\underline{\boxed{\bf{PV=nRT}}}}$

where,

P is pressure

V is volume

n is no. of moles

R is gas constant

T is temperature

• If volume, temperature and same amount of gas is taken then

$\purple{\underline{\boxed{\bf{\dfrac{P}{T}=constant}}}}$

or

$\orange{\underline{\boxed{\bf{\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}}}}}$

where,

P₁ initial pressure

T₁ is initial temperature

P₂ is final pressure

T₂ is final temperature

━━━━━━━━━━━━━━━━━━━━━━━━━

Let the final temperature be T₂

So,

$\longrightarrow\rm{\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}}$

$\longrightarrow\rm{\dfrac{1.5}{300}=\dfrac{1.75}{T_{2}}}$

$\longrightarrow\rm{T_{2}=\dfrac{1.75\times300}{1.5}}$

$\longrightarrow\rm\green{T_{2}=350^{\circ}C}$

Hence, the final temperature is 350°C.

Answer 2

$\large\bf{\underline{\underline{\red{SOLUTION:-}}}}$

GIVEN :-

✴️ A car tyre has air at 1.5 atm at 300K .

✴️ Pressure increases to 1.75 atm at constant volume .

• $\rm{P_{initial}\:=\:1.5\:atm}$

• $\rm{P_{final}\:=\:1.75\:atm}$

• $\rm{T_{initial}\:=\:300\:K}$

To Find :-

• The temperature, i.e $\rm{T_{final}}$ .

FORMULA :-

✴️ We have know that the ideal gas equation, i.e

$\bigstar\:\rm{\underline{\purple{\boxed{PV\:=\:nRT}}}}$

• P = Pressure
• V = Volume
• R = Universal gas constant
• T = Temperature

✪ Here, ‘n’ , ‘ R’ and ‘V’ are constant terms .

$\rm{\implies\:P\:=\:(\dfrac{nR}{V})\:T\:}$

✪ In the above equation, ( nR/V ) = constant = k .

$\rm{\implies\:P\:=\:kT}$

• ‘k’ is the constant term .

$\rm{\implies\:P\:\alpha\:T}$

$\bigstar\:\large\rm{\underline{\underline{\blue{\boxed{\dfrac{P_i}{P_f}\:=\:\dfrac{T_i}{T_f}\:}}}}}$

CALCULATION :-

$\rm{\implies\:T_f\:=\:\dfrac{P_f}{P_i}\:\times{T_i}\:}$

$\rm{\implies\:T_f\:=\:\dfrac{1.75}{1.5}\:\times{300}\:}$

$\rm{\green{\boxed{\implies\:T_f\:=\:350\:K\:}}}$

▼ Therefore, the temperature is 350K .