Question

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

• A. 350${}^0$C
• B. 350K
• C. 300${}^0$C
• D. 300K

Answer 1

The correct option is B 350K

PV = nRT
Since n and R are constant, we get
$\frac{PV}{T}$ = constant
so, $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
Here,
$P_1=1.5atm$
$V_1=V_2$
$T_1=300K$
$P_2=1.75atm$
$T_2=?$
Putting all the values, we get :

$T_2=\left(\frac{P_2{V}_2}{P_1{V}_1}\right)T_1$

$T_2=\frac{1.75}{1.5}\times 300$

$=350K$