Question

can anyone solve this
please post the solution

Answer 1

The pressure exerted by 1212 g of an ideal gas at a temperature toto C in a vessel of volume VVis 11 atm. If the temperature is increased by 10o10o C, the pressure increases by 10%.10%. The molecular weight of the gas is 120.120. We want to determine the volume of the gas.

Let the initial pressure and temperature be P1P1 and T1T1 and let the final pressure and temperature be P2P2 and T2.T2. The volume is constant at V.V.

⇒P1T1=P2T2.⇒P1T1=P2T2.

P1=1P1=1 atm = 101325101325 Pa, P2=1.1P1P2=1.1P1

T1=273.16+tT1=273.16+t K, T2=273.16+t+10=283.16+tT2=273.16+t+10=283.16+tK.

⇒P1273.16+t=1.1P1283.16+t.⇒P1273.16+t=1.1P1283.16+t.

⇒283.16+t=1.1(273.16+t)=300.48+1.1t.⇒283.16+t=1.1(273.16+t)=300.48+1.1t.

⇒0.1t=283.16−300.48=−17.32⇒t=−173.2.⇒0.1t=283.16−300.48=−17.32⇒t=−173.2.

⇒T1=99.96⇒T1=99.96 K.

We have 1212 g of gas with molecular weight 120.120.

⇒⇒ The quantity of gas is n=12120=0.1n=12120=0.1 mole.

By the ideal gas equation, P1V=nRT1⇒V=nRT1P1.P1V=nRT1⇒V=nRT1P1.

R=8.314J/mol.K.R=8.314J/mol.K.

⇒V=0.1×8.314×99.96101325=8.2×10−4⇒V=0.1×8.314×99.96101325=8.2×10−4m3.3.

⇒⇒ The volume of gas is 8.2×10−48.2×10−4 m3.

Pls mark me brainlist