Question

An ideal gas has molar heat capacity at constant pressure $C_P=\frac{5R}{2}$. The gas is kept in a cylindrical vessel fitted with a piston which is free to move. Mass of the frictionless piston is $9kg$. Initial volume of the gas is $0.0027m^3$ and cross section area of the piston is $0.09m^2$. The initial temperature of the gas is 300 $K$. Atmospheric pressure $P_0=1.05\times {10}^{5}N/m^2$. An amount of $2.5\times {10}^{4}J$ of heat energy is supplied to the gas such that piston moves without acceleration, then

• A. Initial pressure of the gas is $1.06\times {10}^{5}N/m^2$
• B. Final temperature of the gas is $1000K$
• C. Final pressure of the gas is $1.06\times {10}^{5}N/m^2$
• D. Work done by gas is ${10}^{4}J$

Answer 1

Answer: (A), (C), (D)

The correct options are
A Initial pressure of the gas is $1.06\times {10}^{5}N/m^2$
C Final pressure of the gas is $1.06\times {10}^{5}N/m^2$
D Work done by gas is ${10}^{4}J$

$PV=nRT$ ...........(1)
Mass of piston is 9 $kg$ & area of piston is $0.09m^2$
So $P=P_0+\frac{9g}{0.09}$
$P=1.06\times {10}^{5}N/m^2$
As mass of piston is not changed & Piston is frictionless So process is isobaric
$P_f=P_i=1.06\times {10}^{5}N/m^2$
$dQ=n{C}_{p}dT$ ............(2)
or $P_2{V}_2-P_1{V}_1=nRdT$ ...........(3)
Dividing (3) by (2)
$\frac{P(V_2-V_1)}{dQ}=\frac{2}{5}$
$\Rightarrow P(V_2-V_1)=\frac{2}{5}\times 2.5\times {10}^4$
$W=P\Delta V={10}^{4}J$ & by equation (1)
$n=\frac{1.06\times {10}^5\times 0.0027}{R\times 300}$
$n=\frac{1.06\times 270}{2500}=0.11448$
or by $dQ=nCpdT$
$\because T_2-T_1=\frac{2.5\times {10}^4}{0.11448\times \frac{5R}{2}}$
On solving $T_2=10782.18K$