Two moles of a gas at 8.21 bar and 300 K are expanded at constant temperature up to 2.73 bar against a constant pressure of 1 bar. How much work (in Latm) is done by the gas? (neglect the sign)

Question

Two moles of a gas at 8.21 bar and 300 K are expanded at constant temperature up to 2.73 bar against a constant pressure of 1 bar. How much work (in Latm) is done by the gas? (neglect the sign) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

V1=(2 × 0.0821 × 300/8.21)=6.0 L V2=(2 × 0.0821 × 300/2.73)=18.0 L Wi r r , i s o t h e r m a l=-Pe x t× Δ V=-1× (.18-6.) =-12 L atm

Q. Two moles of a gas at 8.21 bar and 300 K are expanded at constant temperature up to 2.73 bar against a constant pressure of 1 bar. How much work (in Latm) is done by the gas?

(neglect the sign)

$V_{1} = \frac{2 \times 0.0821 \times 300}{8.21} = 6.0 L$

$V_{2} = \frac{2 \times 0.0821 \times 300}{2.73} = 18.0 L$

$W_{i rr, i so t h er ma l} = - P_{e x t} \times Δ V = - 1 \times (18 - 6)$

$= - 12 L a t m$

Q. Two moles of a gas at 8.21 bar and 300 K are expanded at constant temperature up to 2.73 bar against a constant pressure of 1 bar. How much work (in Latm) is done by the gas? (neglect the sign)

V1​=8.212×0.0821×300​=6.0L V2​=2.732×0.0821×300​=18.0L Wirr,isothermal​=−Pext​×ΔV=−1×(18−6) =−12Latm

Q. Two moles of a gas at 8.21 bar and 300 K are expanded at constant temperature up to 2.73 bar against a constant pressure of 1 bar. How much work (in Latm) is done by the gas?

(neglect the sign)

$V_{1} = \frac{2 \times 0.0821 \times 300}{8.21} = 6.0 L$

$V_{2} = \frac{2 \times 0.0821 \times 300}{2.73} = 18.0 L$

$W_{i rr, i so t h er ma l} = - P_{e x t} \times Δ V = - 1 \times (18 - 6)$

$= - 12 L a t m$