Question

Fixed mass of an ideal gas at $290$K and $1$ atm pressure occupied a volume of $0.2m^2$. Calculate the volume of the gas if pressure and temperature and changed to $1.2159\times {10}^5$ Pa and $253$k respectively.

Answer 1

Man is fixed so number mole is fixed so $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$P_1=1atm$ $V_1=0.2m^2$ $t_1=290k$

$P_2=1.2159\times {10}^{5}pa$ $T=253k$ $V_2?$

$\frac{P_1{V}_1}{T_1}$ $\frac{P_2{V}_2}{T_2}$

$\frac{P_1{V}_1{T}_2}{P_2{T}_1}=V_2$

$V_2=\frac{1.01325\times {10}^5\times 0.2\times 253}{1.2159\times {10}^5\times 290}$

$=0.1459m^3$