Two gases occupy two containers A and B the gas in A, of volume 0.10 m^ 3 , exerts a pressure of 1.40 MPa and that in B of volume 0.15 m^ 3 exerts a pressure 0.7 MPa . The two containers are united by a tube of

Correct answer: B. 0.98 We know that P_ A V_ A =n_ A RT, P_ B V_ B =n_ B RT array ll & and P_ f (V_ A +V_ B )=(n_ A +n_ B ) R T & P_ f (V_ A +V_ B )=P_ A V_ A +P_ B V_ B & P_ f =( P_ A V_ A +P_ B V_ B V_ A +V_ B ) & =(1.4 × 0.1+0.7 × 0.15)/(0.1+0.15) MPa=0.98 MPa array

BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Physics
Chapter: Thermodynamics

BITSAT 2013 · Physics · Thermodynamics

Two gases occupy two containers A and B the gas in A, of volume...

Two gases occupy two containers

A

and

B

the gas in

\mathrm{A},

of volume

0.10 \mathrm{ m}^{3},

exerts a pressure of

1.40 \mathrm{MPa}

and that in

\mathrm{B}

of volume

0.15 \mathrm{ m}^{3}

exerts a pressure

0.7 \mathrm{MPa}

. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)

Options

A0.7
B0.98
C1.4
D210

Correct answer

B. 0.98

Step-by-step solution

We know that

\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}=\mathrm{n}_{\mathrm{A}} \mathrm{RT}, \mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}=\mathrm{n}_{\mathrm{B}} \mathrm{RT}

\begin{array}{ll} & \text{ and } P_{f}\left(V_{A}+V_{B}\right)=\left(n_{A}+n_{B}\right) R T \\ & P_{f}\left(V_{A}+V_{B}\right)=P_{A} V_{A}+P_{B} V_{B} \\ \therefore & P_{f}=\left(\frac{P_{A} V_{A}+P_{B} V_{B}}{V_{A}+V_{B}}\right) \\ & =\frac{1.4 \times 0.1+0.7 \times 0.15}{0.1+0.15} \mathrm{MPa}=0.98 \mathrm{MPa}\end{array}

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